Which prototype does the print() call match here? It doesn’t match any of them! A lack of a matching prototype doesn’t automatically rule out using one of the functions because C++ will try to use standard type conversions to force a match. If, say, the
Some signatures that appear to be different from each other nonetheless can’t coexist. For example, consider these two prototypes:
double cube(double x);
double cube(double & x);
You might think this is a place you could use function overloading because the function signatures appear to be different. But consider things from the compiler’s standpoint. Suppose you have code like this:
cout << cube(x);
The x argument matches both the double x prototype and the double &x prototype. The compiler has no way of knowing which function to use. Therefore, to avoid such confusion, when it checks function signatures, the compiler considers a reference to a type and the type itself to be the same signature.
The function-matching process does discriminate between const and non-const variables. Consider the following prototypes:
void dribble(char * bits); // overloaded
void dribble (const char *cbits); // overloaded
void dabble(char * bits); // not overloaded
void drivel(const char * bits); // not overloaded
Here’s what various function calls would match:
const char p1[20] = "How's the weather?";
char p2[20] = "How's business?";
dribble(p1); // dribble(const char *);
dribble(p2); // dribble(char *);
dabble(p1); // no match
dabble(p2); // dabble(char *);
drivel(p1); // drivel(const char *);
drivel(p2); // drivel(const char *);
The dribble() function has two prototypes—one for const pointers and one for regular pointers—and the compiler selects one or the other, depending on whether the actual argument is const. The dabble() function only matches a call with a non-const argument, but the drivel() function matches calls with either const or non-const arguments. The reason for this difference in behavior between drivel() and dabble() is that it’s valid to assign a non-const value to a const variable, but not vice versa.
Keep in mind that the signature, not the function type, enables function overloading. For example, the following two declarations are incompatible:
long gronk(int n, float m); // same signatures,
double gronk(int n, float m); // hence not allowed
Therefore, C++ doesn’t permit you to overload gronk() in this fashion. You can have different return types, but only if the signatures are also different:
long gronk(int n, float m); // different signatures,
double gronk(float n, float m); // hence allowed
After we discuss templates later in this chapter, we’ll further discuss function matching.
Overloading Reference Parameters
Class designs and the STL often use reference parameters, and it’s useful to know how overloading works with different reference types. Consider the following three prototypes:
void sink(double & r1); // matches modifiable lvalue
void sank(const double & r2); // matches modifiable or const lvalue, rvalue
void sunk(double && r3); // matches rvalue
The lvalue reference parameter r1 matches a modifiable lvalue argument, such as a double variable. The const lvalue reference parameter r2 matches a modifiable lvalue argument, a const lvalue argument, and an rvalue argument, such as the sum of two double values. Finally, the rvalue reference r3 matches an rvalue. Note how r2 can match the same sort of arguments that r1 and r3 match. This raises the question of what happens when you overload a function on these three types of parameters. The answer is that the more exact match is made:
void staff(double & rs); // matches modifiable lvalue
voit staff(const double & rcs); // matches rvalue, const lvalue
void stove(double & r1); // matches modifiable lvalue
void stove(const double & r2); // matches const lvalue
void stove(double && r3); // matches rvalue
This allows you to customize the behavior of a function based on the lvalue, const, or rvalue nature of the argument:
double x = 55.5;
const double y = 32.0;
stove(x); // calls stove(double &)
stove(y); // calls stove(const double &)
stove(x+y); // calls stove(double &&)
If, say, you omit the stove(double &&) function, then stove(x+y) will call the stove(const double &) function instead.
An Overloading Example