What is void (*pfunc[3])()?

Question

I have this code line in my code. what is void (*pfunc[3])() ?

uint32_t Var1 = 1, Var2 = 2, Var3 = 3;
void (*pfunc[3])();
pfunc[0] = Var1;
pfunc[1] = Var2;
pfunc[2] = Var3;

Answer 1

The code snippet does not make sense.

This

void (*pfunc[3])();

is a declaration of an array of three elements of pointers to functions of the type void(). Moreover there is nothing known about the function parameters.

On the other hand. this declaration

uint32_t Var1 = 1, Var2 = 2, Var3 = 3;

declares three objects of the type uint32_t.

So the compiler should issue at least a warning that there is no implicit conversion from the type uint32_t to the type void (* )()for statements like this

pfunc[0] = Var1;

Answer 2

your example does not make too much sense bu

void foo()
{
    printf("foo\n");
}

void bar()
{
    printf("bar\n");
}

void goo()
{
    printf("goo\n");
}



int main()
{
    uint64_t Var1 = (uint64_t)foo, Var2 = (uint64_t)bar, Var3 = (uint64_t)goo;
    void (*pfunc[3])();
    pfunc[0] = (void (*)())Var1;
    pfunc[1] = (void (*)())Var2;
    pfunc[2] = (void (*)())Var3;

    pfunc[0]();
    pfunc[1]();
    pfunc[2]();
}

Var1, Var2, Var3 keep addresses of the functions.

void (*pfunc[3])(); - declares an array of three function pointers

pfunc[2](); - dereferences the functions pointer - ie calls the referenced function

https://godbolt.org/z/VUpfLl

PS uint64_t was used because integer has to have a size at least same as the size of the pointer. My system is 64 bits so the pointers are 64 bits.

Answer 3

Addendum to @VladfromMoscow's answer.

The syntax of the line you ask about can look a bit odd if you're not familiar with that sort of thing. If you re-state it as:

typedef void (*fn_ptr_t)();   
  // fn_ptr_t is a type of pointer, pointing to a function returning void and taking no arguments.

...

fn_ptr_t pfunc[3];
  // an array of three fn_ptr_t's

then it might be a bit clearer. Specifically, your array declaration now looks like a 'normal' array (like int numbers[4] for example).

You could argue that C's weird syntax has now pushed the confusion one step away into the typedef... Try this: Typedef function pointer?