difference between void (func)(void) and void (*func)(void) when declared as a field inside a struct-CodePudding

So I was told that these are pretty much identical to each other

void function1 (void(func)(int), int arg){
    func(arg);
}

void function2 (void(*func)(int), int arg){
    func(arg);
}

When I execute this it throws no errors.

void print_plus_1(int p)
{
    printf("p=%d\n", p 1);
}

void print_plus_2(int p)
{
    printf("p=%d\n", p 2);
}


int main(void)
{
    function1(print_plus_1, 1);
    function2(print_plus_2, 1);

}

However when I initialize the same void (func) (int) vs void (*func) (int) pattern inside a struct

struct foo
{
    void (func1)(int);
    void (*func2)(int);
};


void print_plus_1(int p)
{
    printf("p=%d\n", p 1);
}

void print_plus_2(int p)
{
    printf("p=%d\n", p 2);
}

int main(void)
{
    struct foo f;
    f.func1 = print_plus_1;
    f.func2 = print_plus_2;
    return 0;
}

It throws a compiler error?

error: field 'func1' declared as a function
    void (func1)(int);

Why does void (func)(void) work when written as a callback function but not as a field inside a struct?

CodePudding user response：

Nominally, void (func)(int) declares a function that takes an int and does not return anything, and void (*func)(int) declares a pointer to such a function.

In standard base C, functions are not objects. They cannot be assigned to variables, passed as arguments, or stored in structure members.

However, C 2018 6.7.6.3 8 says:

A declaration of a parameter as “function returning type” shall be adjusted to “pointer to function returning type”, as in 6.3.2.1.

This means that, when you declare a parameter to be a function, it is automatically changed to be declare the parameter to be a pointer to a function. There is nothing in the C standard that says to do this adjustment when declaring a structure member. That is why you may nominally (but not actually) declare a function parameter to be a function but cannot declare a structure member to be a function.