Function pointers with default parameters in C++

Question

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?

C++ does not allow default arguments in function pointer declarations. Your func_ptr3 declaration is ill-formed. So there's nothing to "handle" here. — AnT stands with Russia
– AnT stands with Russia, Commented Apr 7, 2021 at 13:39

Georg Fritzsche · Accepted Answer · 2010-04-05 00:30:52Z

36

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]

answered Apr 5, 2010 at 0:30

Georg Fritzsche

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4 Comments

YeenFei Over a year ago

because default parameter are actually implemented by compiler ?

Georg Fritzsche Over a year ago

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.

Daniel Daranas Over a year ago

That would rule out func_ptr1. What about func_ptr3?

Georg Fritzsche Over a year ago

That also rules out #3 - default arguments are only allowed for function declarations.

Dixit Singla · Accepted Answer · 2013-08-02 16:55:11Z

1

Default argument cannot be provided for pointers to functions.

answered Aug 2, 2013 at 16:55

Dixit Singla

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