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Thinking in C++, 2nd ed. Volume 1

©2000 by Bruce Eckel

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11: References & the Copy-Constructor

References are like constant pointers that are automatically dereferenced by the compiler.

Although references also exist in Pascal, the C++ version was taken from the Algol language. They are essential in C++ to support the syntax of operator overloading (see Chapter 12), but they are also a general convenience to control the way arguments are passed into and out of functions.

This chapter will first look briefly at the differences between pointers in C and C++, then introduce references. But the bulk of the chapter will delve into a rather confusing issue for the new C++ programmer: the copy-constructor, a special constructor (requiring references) that makes a new object from an existing object of the same type. The copy-constructor is used by the compiler to pass and return objects by value into and out of functions.

Finally, the somewhat obscure C++ pointer-to-member feature is illuminated.

Pointers in C++

The most important difference between pointers in C and those in C++ is that C++ is a more strongly typed language. This stands out where void* is concerned. C doesn’t let you casually assign a pointer of one type to another, but it does allow you to accomplish this through a void*. Thus,

bird* b;
rock* r;
void* v;
v = r;
b = v;

Because this “feature” of C allows you to quietly treat any type like any other type, it leaves a big hole in the type system. C++ doesn’t allow this; the compiler gives you an error message, and if you really want to treat one type as another, you must make it explicit, both to the compiler and to the reader, using a cast. (Chapter 3 introduced C++’s improved “explicit” casting syntax.)

References in C++

A reference (&) is like a constant pointer that is automatically dereferenced. It is usually used for function argument lists and function return values. But you can also make a free-standing reference. For example,

//: C11:FreeStandingReferences.cpp
#include <iostream>
using namespace std;

// Ordinary free-standing reference:
int y;
int& r = y;
// When a reference is created, it must 
// be initialized to a live object. 
// However, you can also say:
const int& q = 12;  // (1)
// References are tied to someone else's storage:
int x = 0;          // (2)
int& a = x;         // (3)
int main() {
  cout << "x = " << x << ", a = " << a << endl;
  cout << "x = " << x << ", a = " << a << endl;
} ///:~

In line (1), the compiler allocates a piece of storage, initializes it with the value 12, and ties the reference to that piece of storage. The point is that any reference must be tied to someone else’s piece of storage. When you access a reference, you’re accessing that storage. Thus, if you write lines like (2) and (3), then incrementing a is actually incrementing x, as is shown in main( ). Again, the easiest way to think about a reference is as a fancy pointer. One advantage of this “pointer” is that you never have to wonder whether it’s been initialized (the compiler enforces it) and how to dereference it (the compiler does it).

There are certain rules when using references:

  1. A reference must be initialized when it is created. (Pointers can be initialized at any time.)
  2. Once a reference is initialized to an object, it cannot be changed to refer to another object. (Pointers can be pointed to another object at any time.)
  3. You cannot have NULL references. You must always be able to assume that a reference is connected to a legitimate piece of storage.

References in functions

The most common place you’ll see references is as function arguments and return values. When a reference is used as a function argument, any modification to the reference inside the function will cause changes to the argument outside the function. Of course, you could do the same thing by passing a pointer, but a reference has much cleaner syntax. (You can think of a reference as nothing more than a syntax convenience, if you want.)

If you return a reference from a function, you must take the same care as if you return a pointer from a function. Whatever the reference is connected to shouldn’t go away when the function returns, otherwise you’ll be referring to unknown memory.

Here’s an example:

//: C11:Reference.cpp
// Simple C++ references

int* f(int* x) {
  return x; // Safe, x is outside this scope

int& g(int& x) {
  x++; // Same effect as in f()
  return x; // Safe, outside this scope

int& h() {
  int q;
//!  return q;  // Error
  static int x;
  return x; // Safe, x lives outside this scope

int main() {
  int a = 0;
  f(&a); // Ugly (but explicit)
  g(a);  // Clean (but hidden)
} ///:~

The call to f( ) doesn’t have the convenience and cleanliness of using references, but it’s clear that an address is being passed. In the call to g( ), an address is being passed (via a reference), but you don’t see it.

const references

The reference argument in Reference.cpp works only when the argument is a non-const object. If it is a const object, the function g( ) will not accept the argument, which is actually a good thing, because the function does modify the outside argument. If you know the function will respect the constness of an object, making the argument a const reference will allow the function to be used in all situations. This means that, for built-in types, the function will not modify the argument, and for user-defined types, the function will call only const member functions, and won’t modify any public data members.

The use of const references in function arguments is especially important because your function may receive a temporary object. This might have been created as a return value of another function or explicitly by the user of your function. Temporary objects are always const, so if you don’t use a const reference, that argument won’t be accepted by the compiler. As a very simple example,

//: C11:ConstReferenceArguments.cpp
// Passing references as const

void f(int&) {}
void g(const int&) {}

int main() {
//!  f(1); // Error
} ///:~

The call to f(1) causes a compile-time error because the compiler must first create a reference. It does so by allocating storage for an int, initializing it to one and producing the address to bind to the reference. The storage must be a const because changing it would make no sense – you can never get your hands on it again. With all temporary objects you must make the same assumption: that they’re inaccessible. It’s valuable for the compiler to tell you when you’re changing such data because the result would be lost information.

Pointer references

In C, if you want to modify the contents of the pointer rather than what it points to, your function declaration looks like:

void f(int**);

and you’d have to take the address of the pointer when passing it in:

int i = 47;
int* ip = &i;

With references in C++, the syntax is cleaner. The function argument becomes a reference to a pointer, and you no longer have to take the address of that pointer. Thus,

//: C11:ReferenceToPointer.cpp
#include <iostream>
using namespace std;

void increment(int*& i) { i++; }

int main() {
  int* i = 0;
  cout << "i = " << i << endl;
  cout << "i = " << i << endl;
} ///:~

By running this program, you’ll prove to yourself that the pointer is incremented, not what it points to.

Argument-passing guidelines

Your normal habit when passing an argument to a function should be to pass by const reference. Although at first this may seem like only an efficiency concern (and you normally don’t want to concern yourself with efficiency tuning while you’re designing and assembling your program), there’s more at stake: as you’ll see in the remainder of the chapter, a copy-constructor is required to pass an object by value, and this isn’t always available.

The efficiency savings can be substantial for such a simple habit: to pass an argument by value requires a constructor and destructor call, but if you’re not going to modify the argument then passing by const reference only needs an address pushed on the stack.

In fact, virtually the only time passing an address isn’t preferable is when you’re going to do such damage to an object that passing by value is the only safe approach (rather than modifying the outside object, something the caller doesn’t usually expect). This is the subject of the next section.

The copy-constructor

Now that you understand the basics of the reference in C++, you’re ready to tackle one of the more confusing concepts in the language: the copy-constructor, often called X(X&) (“X of X ref”). This constructor is essential to control passing and returning of user-defined types by value during function calls. It’s so important, in fact, that the compiler will automatically synthesize a copy-constructor if you don’t provide one yourself, as you will see.

Passing & returning by value

To understand the need for the copy-constructor, consider the way C handles passing and returning variables by value during function calls. If you declare a function and make a function call,

int f(int x, char c);
int g = f(a, b);

how does the compiler know how to pass and return those variables? It just knows! The range of the types it must deal with is so small – char, int, float, double, and their variations – that this information is built into the compiler.

If you figure out how to generate assembly code with your compiler and determine the statements generated by the function call to f( ), you’ll get the equivalent of:

push  b
push  a
call  f()
add  sp,4
mov  g, register a

This code has been cleaned up significantly to make it generic; the expressions for b and a will be different depending on whether the variables are global (in which case they will be _b and _a) or local (the compiler will index them off the stack pointer). This is also true for the expression for g. The appearance of the call to f( ) will depend on your name-decoration scheme, and “register a” depends on how the CPU registers are named within your assembler. The logic behind the code, however, will remain the same.

In C and C++, arguments are first pushed on the stack from right to left, then the function call is made. The calling code is responsible for cleaning the arguments off the stack (which accounts for the add sp,4). But notice that to pass the arguments by value, the compiler simply pushes copies on the stack – it knows how big they are and that pushing those arguments makes accurate copies of them.

The return value of f( ) is placed in a register. Again, the compiler knows everything there is to know about the return value type because that type is built into the language, so the compiler can return it by placing it in a register. With the primitive data types in C, the simple act of copying the bits of the value is equivalent to copying the object.

Passing & returning large objects

But now consider user-defined types. If you create a class and you want to pass an object of that class by value, how is the compiler supposed to know what to do? This is not a type built into the compiler; it’s a type you have created.

To investigate this, you can start with a simple structure that is clearly too large to return in registers:

//: C11:PassingBigStructures.cpp
struct Big {
  char buf[100];
  int i;
  long d;
} B, B2;

Big bigfun(Big b) {
  b.i = 100; // Do something to the argument
  return b;

int main() {
  B2 = bigfun(B);
} ///:~

Decoding the assembly output is a little more complicated here because most compilers use “helper” functions instead of putting all functionality inline. In main( ), the call to bigfun( ) starts as you might guess – the entire contents of B is pushed on the stack. (Here, you might see some compilers load registers with the address of the Big and its size, then call a helper function to push the Big onto the stack.)

In the previous code fragment, pushing the arguments onto the stack was all that was required before making the function call. In PassingBigStructures.cpp, however, you’ll see an additional action: the address of B2 is pushed before making the call, even though it’s obviously not an argument. To comprehend what’s going on here, you need to understand the constraints on the compiler when it’s making a function call.

Function-call stack frame

When the compiler generates code for a function call, it first pushes all the arguments on the stack, then makes the call. Inside the function, code is generated to move the stack pointer down even farther to provide storage for the function’s local variables. (“Down” is relative here; your machine may increment or decrement the stack pointer during a push.) But during the assembly-language CALL, the CPU pushes the address in the program code where the function call came from, so the assembly-language RETURN can use that address to return to the calling point. This address is of course sacred, because without it your program will get completely lost. Here’s what the stack frame looks like after the CALL and the allocation of local variable storage in the function:

The code generated for the rest of the function expects the memory to be laid out exactly this way, so that it can carefully pick from the function arguments and local variables without touching the return address. I shall call this block of memory, which is everything used by a function in the process of the function call, the function frame.

You might think it reasonable to try to return values on the stack. The compiler could simply push it, and the function could return an offset to indicate how far down in the stack the return value begins.


The problem occurs because functions in C and C++ support interrupts; that is, the languages are re-entrant. They also support recursive function calls. This means that at any point in the execution of a program an interrupt can occur without breaking the program. Of course, the person who writes the interrupt service routine (ISR) is responsible for saving and restoring all the registers that are used in the ISR, but if the ISR needs to use any memory further down on the stack, this must be a safe thing to do. (You can think of an ISR as an ordinary function with no arguments and void return value that saves and restores the CPU state. An ISR function call is triggered by some hardware event instead of an explicit call from within a program.)

Now imagine what would happen if an ordinary function tried to return values on the stack. You can’t touch any part of the stack that’s above the return address, so the function would have to push the values below the return address. But when the assembly-language RETURN is executed, the stack pointer must be pointing to the return address (or right below it, depending on your machine), so right before the RETURN, the function must move the stack pointer up, thus clearing off all its local variables. If you’re trying to return values on the stack below the return address, you become vulnerable at that moment because an interrupt could come along. The ISR would move the stack pointer down to hold its return address and its local variables and overwrite your return value.

To solve this problem, the caller could be responsible for allocating the extra storage on the stack for the return values before calling the function. However, C was not designed this way, and C++ must be compatible. As you’ll see shortly, the C++ compiler uses a more efficient scheme.

Your next idea might be to return the value in some global data area, but this doesn’t work either. Reentrancy means that any function can be an interrupt routine for any other function, including the same function you’re currently inside. Thus, if you put the return value in a global area, you might return into the same function, which would overwrite that return value. The same logic applies to recursion.

The only safe place to return values is in the registers, so you’re back to the problem of what to do when the registers aren’t large enough to hold the return value. The answer is to push the address of the return value’s destination on the stack as one of the function arguments, and let the function copy the return information directly into the destination. This not only solves all the problems, it’s more efficient. It’s also the reason that, in PassingBigStructures.cpp, the compiler pushes the address of B2 before the call to bigfun( ) in main( ). If you look at the assembly output for bigfun( ), you can see it expects this hidden argument and performs the copy to the destination inside the function.

Bitcopy versus initialization

So far, so good. There’s a workable process for passing and returning large simple structures. But notice that all you have is a way to copy the bits from one place to another, which certainly works fine for the primitive way that C looks at variables. But in C++ objects can be much more sophisticated than a patch of bits; they have meaning. This meaning may not respond well to having its bits copied.

Consider a simple example: a class that knows how many objects of its type exist at any one time. From Chapter 10, you know the way to do this is by including a static data member:

//: C11:HowMany.cpp
// A class that counts its objects
#include <fstream>
#include <string>
using namespace std;
ofstream out("HowMany.out");

class HowMany {
  static int objectCount;
  HowMany() { objectCount++; }
  static void print(const string& msg = "") {
    if(msg.size() != 0) out << msg << ": ";
    out << "objectCount = "
         << objectCount << endl;
  ~HowMany() {

int HowMany::objectCount = 0;

// Pass and return BY VALUE:
HowMany f(HowMany x) {
  x.print("x argument inside f()");
  return x;

int main() {
  HowMany h;
  HowMany::print("after construction of h");
  HowMany h2 = f(h);
  HowMany::print("after call to f()");
} ///:~

The class HowMany contains a static int objectCount and a static member function print( ) to report the value of that objectCount, along with an optional message argument. The constructor increments the count each time an object is created, and the destructor decrements it.

The output, however, is not what you would expect:

after construction of h: objectCount = 1
x argument inside f(): objectCount = 1
~HowMany(): objectCount = 0
after call to f(): objectCount = 0
~HowMany(): objectCount = -1
~HowMany(): objectCount = -2

After h is created, the object count is one, which is fine. But after the call to f( ) you would expect to have an object count of two, because h2 is now in scope as well. Instead, the count is zero, which indicates something has gone horribly wrong. This is confirmed by the fact that the two destructors at the end make the object count go negative, something that should never happen.

Look at the point inside f( ), which occurs after the argument is passed by value. This means the original object h exists outside the function frame, and there’s an additional object inside the function frame, which is the copy that has been passed by value. However, the argument has been passed using C’s primitive notion of bitcopying, whereas the C++ HowMany class requires true initialization to maintain its integrity, so the default bitcopy fails to produce the desired effect.

When the local object goes out of scope at the end of the call to f( ), the destructor is called, which decrements objectCount, so outside the function, objectCount is zero. The creation of h2 is also performed using a bitcopy, so the constructor isn’t called there either, and when h and h2 go out of scope, their destructors cause the negative values of objectCount.


The problem occurs because the compiler makes an assumption about how to create a new object from an existing object. When you pass an object by value, you create a new object, the passed object inside the function frame, from an existing object, the original object outside the function frame. This is also often true when returning an object from a function. In the expression

HowMany h2 = f(h);

h2, a previously unconstructed object, is created from the return value of f( ), so again a new object is created from an existing one.

The compiler’s assumption is that you want to perform this creation using a bitcopy, and in many cases this may work fine, but in HowMany it doesn’t fly because the meaning of initialization goes beyond simply copying. Another common example occurs if the class contains pointers – what do they point to, and should you copy them or should they be connected to some new piece of memory?

Fortunately, you can intervene in this process and prevent the compiler from doing a bitcopy. You do this by defining your own function to be used whenever the compiler needs to make a new object from an existing object. Logically enough, you’re making a new object, so this function is a constructor, and also logically enough, the single argument to this constructor has to do with the object you’re constructing from. But that object can’t be passed into the constructor by value because you’re trying to define the function that handles passing by value, and syntactically it doesn’t make sense to pass a pointer because, after all, you’re creating the new object from an existing object. Here, references come to the rescue, so you take the reference of the source object. This function is called the copy-constructor and is often referred to as X(X&), which is its appearance for a class called X.

If you create a copy-constructor, the compiler will not perform a bitcopy when creating a new object from an existing one. It will always call your copy-constructor. So, if you don’t create a copy-constructor, the compiler will do something sensible, but you have the choice of taking over complete control of the process.

Now it’s possible to fix the problem in HowMany.cpp:

//: C11:HowMany2.cpp
// The copy-constructor
#include <fstream>
#include <string>
using namespace std;
ofstream out("HowMany2.out");

class HowMany2 {
  string name; // Object identifier
  static int objectCount;
  HowMany2(const string& id = "") : name(id) {
  ~HowMany2() {
  // The copy-constructor:
  HowMany2(const HowMany2& h) : name(h.name) {
    name += " copy";
    print("HowMany2(const HowMany2&)");
  void print(const string& msg = "") const {
    if(msg.size() != 0) 
      out << msg << endl;
    out << '\t' << name << ": "
        << "objectCount = "
        << objectCount << endl;

int HowMany2::objectCount = 0;

// Pass and return BY VALUE:
HowMany2 f(HowMany2 x) {
  x.print("x argument inside f()");
  out << "Returning from f()" << endl;
  return x;

int main() {
  HowMany2 h("h");
  out << "Entering f()" << endl;
  HowMany2 h2 = f(h);
  h2.print("h2 after call to f()");
  out << "Call f(), no return value" << endl;
  out << "After call to f()" << endl;
} ///:~

There are a number of new twists thrown in here so you can get a better idea of what’s happening. First, the string name acts as an object identifier when information about that object is printed. In the constructor, you can put an identifier string (usually the name of the object) that is copied to name using the string constructor. The default = "" creates an empty string. The constructor increments the objectCount as before, and the destructor decrements it.

Next is the copy-constructor, HowMany2(const HowMany2&). The copy-constructor can create a new object only from an existing one, so the existing object’s name is copied to name, followed by the word “copy” so you can see where it came from. If you look closely, you’ll see that the call name(h.name) in the constructor initializer list is actually calling the string copy-constructor.

Inside the copy-constructor, the object count is incremented just as it is inside the normal constructor. This means you’ll now get an accurate object count when passing and returning by value.

The print( ) function has been modified to print out a message, the object identifier, and the object count. It must now access the name data of a particular object, so it can no longer be a static member function.

Inside main( ), you can see that a second call to f( ) has been added. However, this call uses the common C approach of ignoring the return value. But now that you know how the value is returned (that is, code inside the function handles the return process, putting the result in a destination whose address is passed as a hidden argument), you might wonder what happens when the return value is ignored. The output of the program will throw some illumination on this.

Before showing the output, here’s a little program that uses iostreams to add line numbers to any file:

//: C11:Linenum.cpp
//{T} Linenum.cpp
// Add line numbers
#include "../require.h"
#include <vector>
#include <string>
#include <fstream>
#include <iostream>
#include <cmath>
using namespace std;

int main(int argc, char* argv[]) {
  requireArgs(argc, 1, "Usage: linenum file\n"
    "Adds line numbers to file");
  ifstream in(argv[1]);
  assure(in, argv[1]);
  string line;
  vector<string> lines;
  while(getline(in, line)) // Read in entire file
  if(lines.size() == 0) return 0;
  int num = 0;
  // Number of lines in file determines width:
  const int width = 
    int(log10((double)lines.size())) + 1;
  for(int i = 0; i < lines.size(); i++) {
    cout.setf(ios::right, ios::adjustfield);
    cout << ++num << ") " << lines[i] << endl;
} ///:~

The entire file is read into a vector<string>, using the same code that you’ve seen earlier in the book. When printing the line numbers, we’d like all the lines to be aligned with each other, and this requires adjusting for the number of lines in the file so that the width allowed for the line numbers is consistent. We can easily determine the number of lines using vector::size( ), but what we really need to know is whether there are more than 10 lines, 100 lines, 1,000 lines, etc. If you take the logarithm, base 10, of the number of lines in the file, truncate it to an int and add one to the value, you’ll find out the maximum width that your line count will be.

You’ll notice a couple of strange calls inside the for loop: setf( ) and width( ). These are ostream calls that allow you to control, in this case, the justification and width of the output. However, they must be called each time a line is output and that is why they are inside the for loop. Volume 2 of this book has an entire chapter explaining iostreams that will tell you more about these calls as well as other ways to control iostreams.

When Linenum.cpp is applied to HowMany2.out, the result is

 1) HowMany2()
 2)   h: objectCount = 1
 3) Entering f()
 4) HowMany2(const HowMany2&)
 5)   h copy: objectCount = 2
 6) x argument inside f()
 7)   h copy: objectCount = 2
 8) Returning from f()
 9) HowMany2(const HowMany2&)
10)   h copy copy: objectCount = 3
11) ~HowMany2()
12)   h copy: objectCount = 2
13) h2 after call to f()
14)   h copy copy: objectCount = 2
15) Call f(), no return value
16) HowMany2(const HowMany2&)
17)   h copy: objectCount = 3
18) x argument inside f()
19)   h copy: objectCount = 3
20) Returning from f()
21) HowMany2(const HowMany2&)
22)   h copy copy: objectCount = 4
23) ~HowMany2()
24)   h copy: objectCount = 3
25) ~HowMany2()
26)   h copy copy: objectCount = 2
27) After call to f()
28) ~HowMany2()
29)   h copy copy: objectCount = 1
30) ~HowMany2()
31)   h: objectCount = 0

As you would expect, the first thing that happens is that the normal constructor is called for h, which increments the object count to one. But then, as f( ) is entered, the copy-constructor is quietly called by the compiler to perform the pass-by-value. A new object is created, which is the copy of h (thus the name “h copy”) inside the function frame of f( ), so the object count becomes two, courtesy of the copy-constructor.

Line eight indicates the beginning of the return from f( ). But before the local variable “h copy” can be destroyed (it goes out of scope at the end of the function), it must be copied into the return value, which happens to be h2. A previously unconstructed object (h2) is created from an existing object (the local variable inside f( )), so of course the copy-constructor is used again in line nine. Now the name becomes “h copy copy” for h2’s identifier because it’s being copied from the copy that is the local object inside f( ). After the object is returned, but before the function ends, the object count becomes temporarily three, but then the local object “h copy” is destroyed. After the call to f( ) completes in line 13, there are only two objects, h and h2, and you can see that h2 did indeed end up as “h copy copy.”

Temporary objects

Line 15 begins the call to f(h), this time ignoring the return value. You can see in line 16 that the copy-constructor is called just as before to pass the argument in. And also, as before, line 21 shows the copy-constructor is called for the return value. But the copy-constructor must have an address to work on as its destination (a this pointer). Where does this address come from?

It turns out the compiler can create a temporary object whenever it needs one to properly evaluate an expression. In this case it creates one you don’t even see to act as the destination for the ignored return value of f( ). The lifetime of this temporary object is as short as possible so the landscape doesn’t get cluttered up with temporaries waiting to be destroyed and taking up valuable resources. In some cases, the temporary might immediately be passed to another function, but in this case it isn’t needed after the function call, so as soon as the function call ends by calling the destructor for the local object (lines 23 and 24), the temporary object is destroyed (lines 25 and 26).

Finally, in lines 28-31, the h2 object is destroyed, followed by h, and the object count goes correctly back to zero.

Default copy-constructor

Because the copy-constructor implements pass and return by value, it’s important that the compiler creates one for you in the case of simple structures – effectively, the same thing it does in C. However, all you’ve seen so far is the default primitive behavior: a bitcopy.

When more complex types are involved, the C++ compiler will still automatically create a copy-constructor if you don’t make one. Again, however, a bitcopy doesn’t make sense, because it doesn’t necessarily implement the proper meaning.

Here’s an example to show the more intelligent approach the compiler takes. Suppose you create a new class composed of objects of several existing classes. This is called, appropriately enough, composition, and it’s one of the ways you can make new classes from existing classes. Now take the role of a naive user who’s trying to solve a problem quickly by creating a new class this way. You don’t know about copy-constructors, so you don’t create one. The example demonstrates what the compiler does while creating the default copy-constructor for your new class:

//: C11:DefaultCopyConstructor.cpp
// Automatic creation of the copy-constructor
#include <iostream>
#include <string>
using namespace std;

class WithCC { // With copy-constructor
  // Explicit default constructor required:
  WithCC() {}
  WithCC(const WithCC&) {
    cout << "WithCC(WithCC&)" << endl;

class WoCC { // Without copy-constructor
  string id;
  WoCC(const string& ident = "") : id(ident) {}
  void print(const string& msg = "") const {
    if(msg.size() != 0) cout << msg << ": ";
    cout << id << endl;

class Composite {
  WithCC withcc; // Embedded objects
  WoCC wocc;
  Composite() : wocc("Composite()") {}
  void print(const string& msg = "") const {

int main() {
  Composite c;
  c.print("Contents of c");
  cout << "Calling Composite copy-constructor"
       << endl;
  Composite c2 = c;  // Calls copy-constructor
  c2.print("Contents of c2");
} ///:~

The class WithCC contains a copy-constructor, which simply announces that it has been called, and this brings up an interesting issue. In the class Composite, an object of WithCC is created using a default constructor. If there were no constructors at all in WithCC, the compiler would automatically create a default constructor, which would do nothing in this case. However, if you add a copy-constructor, you’ve told the compiler you’re going to handle constructor creation, so it no longer creates a default constructor for you and will complain unless you explicitly create a default constructor as was done for WithCC.

The class WoCC has no copy-constructor, but its constructor will store a message in an internal string that can be printed out using print( ). This constructor is explicitly called in Composite’s constructor initializer list (briefly introduced in Chapter 8 and covered fully in Chapter 14). The reason for this becomes apparent later.

The class Composite has member objects of both WithCC and WoCC (note the embedded object wocc is initialized in the constructor-initializer list, as it must be), and no explicitly defined copy-constructor. However, in main( ) an object is created using the copy-constructor in the definition:

Composite c2 = c;

The copy-constructor for Composite is created automatically by the compiler, and the output of the program reveals the way that it is created:

Contents of c: Composite()
Calling Composite copy-constructor
Contents of c2: Composite()

To create a copy-constructor for a class that uses composition (and inheritance, which is introduced in Chapter 14), the compiler recursively calls the copy-constructors for all the member objects and base classes. That is, if the member object also contains another object, its copy-constructor is also called. So in this case, the compiler calls the copy-constructor for WithCC. The output shows this constructor being called. Because WoCC has no copy-constructor, the compiler creates one for it that just performs a bitcopy, and calls that inside the Composite copy-constructor. The call to Composite::print( ) in main shows that this happens because the contents of c2.wocc are identical to the contents of c.wocc. The process the compiler goes through to synthesize a copy-constructor is called memberwise initialization.

It’s always best to create your own copy-constructor instead of letting the compiler do it for you. This guarantees that it will be under your control.

Alternatives to copy-construction

At this point your head may be swimming, and you might be wondering how you could have possibly written a working class without knowing about the copy-constructor. But remember: You need a copy-constructor only if you’re going to pass an object of your class by value. If that never happens, you don’t need a copy-constructor.

Preventing pass-by-value

“But,” you say, “if I don’t make a copy-constructor, the compiler will create one for me. So how do I know that an object will never be passed by value?”

There’s a simple technique for preventing pass-by-value: declare a private copy-constructor. You don’t even need to create a definition, unless one of your member functions or a friend function needs to perform a pass-by-value. If the user tries to pass or return the object by value, the compiler will produce an error message because the copy-constructor is private. It can no longer create a default copy-constructor because you’ve explicitly stated that you’re taking over that job.

Here’s an example:

//: C11:NoCopyConstruction.cpp
// Preventing copy-construction

class NoCC {
  int i;
  NoCC(const NoCC&); // No definition
  NoCC(int ii = 0) : i(ii) {}

void f(NoCC);

int main() {
  NoCC n;
//! f(n); // Error: copy-constructor called
//! NoCC n2 = n; // Error: c-c called
//! NoCC n3(n); // Error: c-c called
} ///:~

Notice the use of the more general form

NoCC(const NoCC&);

using the const.

Functions that modify outside objects

Reference syntax is nicer to use than pointer syntax, yet it clouds the meaning for the reader. For example, in the iostreams library one overloaded version of the get( ) function takes a char& as an argument, and the whole point of the function is to modify its argument by inserting the result of the get( ). However, when you read code using this function it’s not immediately obvious to you that the outside object is being modified:

char c;

Instead, the function call looks like a pass-by-value, which suggests the outside object is not modified.

Because of this, it’s probably safer from a code maintenance standpoint to use pointers when you’re passing the address of an argument to modify. If you always pass addresses as const references except when you intend to modify the outside object via the address, where you pass by non-const pointer, then your code is far easier for the reader to follow.

Pointers to members

A pointer is a variable that holds the address of some location. You can change what a pointer selects at runtime, and the destination of the pointer can be either data or a function. The C++ pointer-to-member follows this same concept, except that what it selects is a location inside a class. The dilemma here is that a pointer needs an address, but there is no “address” inside a class; selecting a member of a class means offsetting into that class. You can’t produce an actual address until you combine that offset with the starting address of a particular object. The syntax of pointers to members requires that you select an object at the same time you’re dereferencing the pointer to member.

To understand this syntax, consider a simple structure, with a pointer sp and an object so for this structure. You can select members with the syntax shown:

//: C11:SimpleStructure.cpp
struct Simple { int a; };
int main() {
  Simple so, *sp = &so;
} ///:~

Now suppose you have an ordinary pointer to an integer, ip. To access what ip is pointing to, you dereference the pointer with a ‘*’:

*ip = 4;

Finally, consider what happens if you have a pointer that happens to point to something inside a class object, even if it does in fact represent an offset into the object. To access what it’s pointing at, you must dereference it with *. But it’s an offset into an object, so you must also refer to that particular object. Thus, the * is combined with the object dereference. So the new syntax becomes –>* for a pointer to an object, and .* for the object or a reference, like this:

objectPointer->*pointerToMember = 47;
object.*pointerToMember = 47;

Now, what is the syntax for defining pointerToMember? Like any pointer, you have to say what type it’s pointing at, and you use a * in the definition. The only difference is that you must say what class of objects this pointer-to-member is used with. Of course, this is accomplished with the name of the class and the scope resolution operator. Thus,

int ObjectClass::*pointerToMember;

defines a pointer-to-member variable called pointerToMember that points to any int inside ObjectClass. You can also initialize the pointer-to-member when you define it (or at any other time):

int ObjectClass::*pointerToMember = &ObjectClass::a;

There is actually no “address” of ObjectClass::a because you’re just referring to the class and not an object of that class. Thus, &ObjectClass::a can be used only as pointer-to-member syntax.

Here’s an example that shows how to create and use pointers to data members:

//: C11:PointerToMemberData.cpp
#include <iostream>
using namespace std;

class Data {
  int a, b, c; 
  void print() const {
    cout << "a = " << a << ", b = " << b
         << ", c = " << c << endl;

int main() {
  Data d, *dp = &d;
  int Data::*pmInt = &Data::a;
  dp->*pmInt = 47;
  pmInt = &Data::b;
  d.*pmInt = 48;
  pmInt = &Data::c;
  dp->*pmInt = 49;
} ///:~

Obviously, these are too awkward to use anywhere except for special cases (which is exactly what they were intended for).

Also, pointers to members are quite limited: they can be assigned only to a specific location inside a class. You could not, for example, increment or compare them as you can with ordinary pointers.


A similar exercise produces the pointer-to-member syntax for member functions. A pointer to a function (introduced at the end of Chapter 3) is defined like this:

int (*fp)(float);

The parentheses around (*fp) are necessary to force the compiler to evaluate the definition properly. Without them this would appear to be a function that returns an int*.

Parentheses also play an important role when defining and using pointers to member functions. If you have a function inside a class, you define a pointer to that member function by inserting the class name and scope resolution operator into an ordinary function pointer definition:

//: C11:PmemFunDefinition.cpp
class Simple2 { 
  int f(float) const { return 1; }
int (Simple2::*fp)(float) const;
int (Simple2::*fp2)(float) const = &Simple2::f;
int main() {
  fp = &Simple2::f;
} ///:~

In the definition for fp2 you can see that a pointer to member function can also be initialized when it is created, or at any other time. Unlike non-member functions, the & is not optional when taking the address of a member function. However, you can give the function identifier without an argument list, because overload resolution can be determined by the type of the pointer to member.

An example

The value of a pointer is that you can change what it points to at runtime, which provides an important flexibility in your programming because through a pointer you can select or change behavior at runtime. A pointer-to-member is no different; it allows you to choose a member at runtime. Typically, your classes will only have member functions publicly visible (data members are usually considered part of the underlying implementation), so the following example selects member functions at runtime.

//: C11:PointerToMemberFunction.cpp
#include <iostream>
using namespace std;

class Widget {
  void f(int) const { cout << "Widget::f()\n"; }
  void g(int) const { cout << "Widget::g()\n"; }
  void h(int) const { cout << "Widget::h()\n"; }
  void i(int) const { cout << "Widget::i()\n"; }

int main() {
  Widget w;
  Widget* wp = &w;
  void (Widget::*pmem)(int) const = &Widget::h;
} ///:~

Of course, it isn’t particularly reasonable to expect the casual user to create such complicated expressions. If the user must directly manipulate a pointer-to-member, then a typedef is in order. To really clean things up, you can use the pointer-to-member as part of the internal implementation mechanism. Here’s the preceding example using a pointer-to-member inside the class. All the user needs to do is pass a number in to select a function.[48]

//: C11:PointerToMemberFunction2.cpp
#include <iostream>
using namespace std;

class Widget {
  void f(int) const { cout << "Widget::f()\n"; }
  void g(int) const { cout << "Widget::g()\n"; }
  void h(int) const { cout << "Widget::h()\n"; }
  void i(int) const { cout << "Widget::i()\n"; }
  enum { cnt = 4 };
  void (Widget::*fptr[cnt])(int) const;
  Widget() {
    fptr[0] = &Widget::f; // Full spec required
    fptr[1] = &Widget::g;
    fptr[2] = &Widget::h;
    fptr[3] = &Widget::i;
  void select(int i, int j) {
    if(i < 0 || i >= cnt) return;
  int count() { return cnt; }

int main() {
  Widget w;
  for(int i = 0; i < w.count(); i++)
    w.select(i, 47);
} ///:~

In the class interface and in main( ), you can see that the entire implementation, including the functions, has been hidden away. The code must even ask for the count( ) of functions. This way, the class implementer can change the quantity of functions in the underlying implementation without affecting the code where the class is used.

The initialization of the pointers-to-members in the constructor may seem overspecified. Shouldn’t you be able to say

fptr[1] = &g;

because the name g occurs in the member function, which is automatically in the scope of the class? The problem is this doesn’t conform to the pointer-to-member syntax, which is required so everyone, especially the compiler, can figure out what’s going on. Similarly, when the pointer-to-member is dereferenced, it seems like


is also over-specified; this looks redundant. Again, the syntax requires that a pointer-to-member always be bound to an object when it is dereferenced.


Pointers in C++ are almost identical to pointers in C, which is good. Otherwise, a lot of C code wouldn’t compile properly under C++. The only compile-time errors you will produce occur with dangerous assignments. If these are in fact what are intended, the compile-time errors can be removed with a simple (and explicit!) cast.

C++ also adds the reference from Algol and Pascal, which is like a constant pointer that is automatically dereferenced by the compiler. A reference holds an address, but you treat it like an object. References are essential for clean syntax with operator overloading (the subject of the next chapter), but they also add syntactic convenience for passing and returning objects for ordinary functions.

The copy-constructor takes a reference to an existing object of the same type as its argument, and it is used to create a new object from an existing one. The compiler automatically calls the copy-constructor when you pass or return an object by value. Although the compiler will automatically create a copy-constructor for you, if you think one will be needed for your class, you should always define it yourself to ensure that the proper behavior occurs. If you don’t want the object passed or returned by value, you should create a private copy-constructor.

Pointers-to-members have the same functionality as ordinary pointers: You can choose a particular region of storage (data or function) at runtime. Pointers-to-members just happen to work with class members instead of with global data or functions. You get the programming flexibility that allows you to change behavior at runtime.


Solutions to selected exercises can be found in the electronic document The Thinking in C++ Annotated Solution Guide, available for a small fee from www.BruceEckel.com.

  1. Turn the “bird & rock” code fragment at the beginning of this chapter into a C program (using structs for the data types), and show that it compiles. Now try to compile it with the C++ compiler and see what happens.
  2. Take the code fragments in the beginning of the section titled “References in C++” and put them into a main( ). Add statements to print output so that you can prove to yourself that references are like pointers that are automatically dereferenced.
  3. Write a program in which you try to (1) Create a reference that is not initialized when it is created. (2) Change a reference to refer to another object after it is initialized. (3) Create a NULL reference.
  4. Write a function that takes a pointer argument, modifies what the pointer points to, and then returns the destination of the pointer as a reference.
  5. Create a class with some member functions, and make that the object that is pointed to by the argument of Exercise 4. Make the pointer a const and make some of the member functions const and prove that you can only call the const member functions inside your function. Make the argument to your function a reference instead of a pointer.
  6. Take the code fragments at the beginning of the section titled “Pointer references” and turn them into a program.
  7. Create a function that takes an argument of a reference to a pointer to a pointer and modifies that argument. In main( ), call the function.
  8. Create a function that takes a char& argument and modifies that argument. In main( ), print out a char variable, call your function for that variable, and print it out again to prove to yourself that it has been changed. How does this affect program readability?
  9. Write a class that has a const member function and a non-const member function. Write three functions that take an object of that class as an argument; the first takes it by value, the second by reference, and the third by const reference. Inside the functions, try to call both member functions of your class and explain the results.
  10. (Somewhat challenging) Write a simple function that takes an int as an argument, increments the value, and returns it. In main( ), call your function. Now discover how your compiler generates assembly code and trace through the assembly statements so that you understand how arguments are passed and returned, and how local variables are indexed off the stack.
  11. Write a function that takes as its arguments a char, int, float, and double. Generate assembly code with your compiler and find the statements that push the arguments on the stack before a function call.
  12. Write a function that returns a double. Generate assembly code and determine how the value is returned.
  13. Produce assembly code for PassingBigStructures.cpp. Trace through and demystify the way your compiler generates code to pass and return large structures.
  14. Write a simple recursive function that decrements its argument and returns zero if the argument becomes zero, otherwise it calls itself. Generate assembly code for this function and explain how the way that the assembly code is created by the compiler supports recursion.
  15. Write code to prove that the compiler automatically synthesizes a copy-constructor if you don’t create one yourself. Prove that the synthesized copy-constructor performs a bitcopy of primitive types and calls the copy-constructor of user-defined types.
  16. Write a class with a copy-constructor that announces itself to cout. Now create a function that passes an object of your new class in by value and another one that creates a local object of your new class and returns it by value. Call these functions to prove to yourself that the copy-constructor is indeed quietly called when passing and returning objects by value.
  17. Create a class that contains a double*. The constructor initializes the double* by calling new double and assigning a value to the resulting storage from the constructor argument. The destructor prints the value that’s pointed to, assigns that value to -1, calls delete for the storage, and then sets the pointer to zero. Now create a function that takes an object of your class by value, and call this function in main( ). What happens? Fix the problem by writing a copy-constructor.
  18. Create a class with a constructor that looks like a copy-constructor, but that has an extra argument with a default value. Show that this is still used as the copy-constructor.
  19. Create a class with a copy-constructor that announces itself. Make a second class containing a member object of the first class, but do not create a copy-constructor. Show that the synthesized copy-constructor in the second class automatically calls the copy-constructor of the first class.
  20. Create a very simple class, and a function that returns an object of that class by value. Create a second function that takes a reference to an object of your class. Call the first function as the argument of the second function, and demonstrate that the second function must use a const reference as its argument.
  21. Create a simple class without a copy-constructor, and a simple function that takes an object of that class by value. Now change your class by adding a private declaration (only) for the copy-constructor. Explain what happens when your function is compiled.
  22. This exercise creates an alternative to using the copy-constructor. Create a class X and declare (but don’t define) a private copy-constructor. Make a public clone( ) function as a const member function that returns a copy of the object that is created using new. Now write a function that takes as an argument a const X& and clones a local copy that can be modified. The drawback to this approach is that you are responsible for explicitly destroying the cloned object (using delete) when you’re done with it.
  23. Explain what’s wrong with both Mem.cpp and MemTest.cpp from Chapter 7. Fix the problem.
  24. Create a class containing a double and a print( ) function that prints the double. In main( ), create pointers to members for both the data member and the function in your class. Create an object of your class and a pointer to that object, and manipulate both class elements via your pointers to members, using both the object and the pointer to the object.
  25. Create a class containing an array of int. Can you index through this array using a pointer to member?
  26. Modify PmemFunDefinition.cpp by adding an overloaded member function f( ) (you can determine the argument list that causes the overload). Now make a second pointer to member, assign it to the overloaded version of f( ), and call the function through that pointer. How does the overload resolution happen in this case?
  27. Start with FunctionTable.cpp from Chapter 3. Create a class that contains a vector of pointers to functions, with add( ) and remove( ) member functions to add and remove pointers to functions. Add a run( ) function that moves through the vector and calls all of the functions.
  28. Modify the above Exercise 27 so that it works with pointers to member functions instead.

[48] Thanks to Owen Mortensen for this example

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Last Update:09/27/2001