Passing a function into another function in C

Question

I want to pass a function into another function, but I have confused myself so much over the syntax and what is actually happening. This is the code I am running:

#include <stdio.h>

int func(int a){
    printf("%d\n", a);
}

void another_func(int p_func(int)){
int (*funcptr)(int) = p_func;
funcptr(7);
}

int main(void){

    another_func(func);

    return 0;
}

However, I have found out that I can put how many asterix I want before func when calling it in main, for example:

another_func(*******************func);

I can do the same when assigning the function to the function pointer, for example:

int (*funcptr)(int) = ***********p_func;

I am though forbidden to put an ampersand in front of p_func when assigning it to the function pointer. Also it doesn't seem to matter if I write:

void another_func(int (*p_func)(int)){

Can my plead for help wake some of the legendary C gods from their slumber who can make a detailed explanation? I really don't get how this works, what is passed etc.

Vlad from Moscow · Accepted Answer · 2017-01-18T00:05:56.940

2

According to the C Standard (6.3.2.1 Lvalues, arrays, and function designators)

4 A function designator is an expression that has function type. Except when it is the operand of the sizeof operator65) or the unary & operator, a function designator with type ‘‘function returning type’’ is converted to an expression that has type ‘‘pointer to function returning type’’

On the other hand (6.5.3.2 Address and indirection operators)

4 The unary * operator denotes indirection. If the operand points to a function, the result is a function designator;

Thus you can place as many asterisks before a function designator as the compiler will allow according to its own limits.:)

And you may use operator & with a function designator. For example

#include <stdio.h>

int f( int x )
{
    return x;
}

int main(void) 
{
    int ( *fp )( int ) = &f;

    printf( "%d\n", fp( 10 ) );

    return 0;
}

Take into account that a function parameter declared as a function is adjusted to pointer to the function type.

Thus these two function declarations declare the same one function and can be both present in a compilation unit (though the function shall be defined once if it is not inline)

void another_func(int p_func(int));
void another_func(int ( *p_func )(int));

edited Jan 18 '17 at 00:05

answered Jan 17 '17 at 23:57

Vlad from Moscow

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So in essence function expressions are always converted to function designators, no matter what? – Noob Doob Jan 18 '17 at 00:04
@NoobDoob A function designator is always converted to function pointer with the rare exceptions listed in the presented quote from the C Standard. – Vlad from Moscow Jan 18 '17 at 00:07
So when I pass the function to another_func it automaticly becomes a pointer thus placing the ampersand before as I said illegal? – Donkey King Jan 18 '17 at 00:10
@DonkeyKing Yes, the argument is implicitly converted to a function pointer. – Vlad from Moscow Jan 18 '17 at 00:11
@DonkeyKing In this regard functions and arrays behave similarly. – Vlad from Moscow Jan 18 '17 at 00:12
But when I put the asterix before the assignation you said it would become a function designator, what exactly is that? – Donkey King Jan 18 '17 at 00:17
@DonkeyKing Yes but at once it is converted to a pointer. – Vlad from Moscow Jan 18 '17 at 00:18

score 0 · Answer 2 · answered Jan 17 '17 at 23:50

0

"I have found out that I can put how many asterix I want before func when calling it in main." I don't want to offend you, but I suggest you to read something about pointers. An example about what you're looking for could be:

int foo(int (*foo2)(void))
{
    foo2();
}

int foo2(void)
{
    return rand();
}

int main(void)
{
    foo(foo2);
}

answered Jan 17 '17 at 23:50

Modfoo

97
9

There are tons of things written about pointers. Elaborate what you mean that I don't understand about pointers. – Donkey King Jan 18 '17 at 00:02

Passing a function into another function in C

2 Answers2