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162

answers:

1

How does C++ handle function pointers in relation to functions with defaulted parameters?

If I have:

void foo(int i, float f = 0.0f);
void bar(int i, float f);


void (*func_ptr1)(int);
void (*func_ptr2)(int, float);
void (*func_ptr3)(int, float = 10.0f);

Which function pointers can I use in relation to which function?

+22  A: 

Both foo() and bar() can only be assigned to func_ptr2.

§8.3.6/2:

A default argument is not part of the type of a function. [Example:

int f(int = 0);

void h() {
    int j = f(1);
    int k = f(); // OK, means f(0)
}

int (*p1)(int) = &f; 
int (*p2)() = &f; // error: type mismatch

--end example]

Georg Fritzsche
because default parameter are actually implemented by compiler ?
YeenFei
Because default arguments are not taken into consideration for the type, i.e. `int f(int)` and `int g(int=0)` have the same type.
Georg Fritzsche
That would rule out func_ptr1. What about func_ptr3?
Daniel Daranas
That also rules out #3 - default arguments are only allowed for function declarations.
Georg Fritzsche