Thinking in C++ - Volume 1

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) {
  (*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
  g(1);
} ///:~

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;
f(&ip);

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;
  increment(i);
  cout << "i = " << i << endl;
} ///:~

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

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.

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.

Re-entrancy

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;
public:
  HowMany() { objectCount++; }
  static void print(const string& msg = "") {
    if(msg.size() != 0) out << msg << ": ";
    out << "objectCount = "
         << objectCount << endl;
  }
  ~HowMany() {
    objectCount--;
    print("~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;
public:
  HowMany2(const string& id = "") : name(id) {
    ++objectCount;
    print("HowMany2()");
  }
  ~HowMany2() {
    --objectCount;
    print("~HowMany2()");
  }
  // The copy-constructor:
  HowMany2(const HowMany2& h) : name(h.name) {
    name += " copy";
    ++objectCount;
    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;
  f(h);
  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
    lines.push_back(line);
  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.width(width);
    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.

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
public:
  // Explicit default constructor required:
  WithCC() {}
  WithCC(const WithCC&) {
    cout << "WithCC(WithCC&)" << endl;
  }
};
 
class WoCC { // Without copy-constructor
  string id;
public:
  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;
public:
  Composite() : wocc("Composite()") {}
  void print(const string& msg = "") const {
    wocc.print(msg);
  }
};
 
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'sconstructor 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
WithCC(WithCC&)
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.

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
public:
  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;
cin.get(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.

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 { 
public: 
  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 {
public:
  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;
  (w.*pmem)(1);
  (wp->*pmem)(2);
} ///:~

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;
public:
  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;
    (this->*fptr[i])(j);
  }
  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

(this->*fptr[i])(j);

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.