C++/C function pointers that return void*

I'm trying to call a function that takes an argument, void(*)(void*, int, const char*), but I cannot figure out how to pass those arguments to the function.

Example:

void ptr(int);
int function(int, int, void(*)(int));

I am trying to call the function like this:

function(20, 20, ptr(20));

Is this possible?

Solution 1:

You are doing one thing incorrectly - you are trying to invoke your 'ptr' function before invoking 'function'. What you were supposed to do is to pass just a pointer to 'ptr' and invoke 'ptr' using passed pointer from 'function' like that:

void ptr(int x)
{
    printf("from ptr [%d]\n", x);
}

int function(int a, int b , void (*func)(int) )
{
    printf( "from function a=[%d] b=[%d]\n", a, b );
    func(a); // you must invoke function here

    return 123;
}


void main()
{
    function( 10, 2, &ptr );
    // or
    function( 20, 2, ptr );
}

which gives:

from function a=[10] b=[2]
from ptr [10]
from function a=[20] b=[2]
from ptr [20]

which is what you wanted

for

function(20, 20, ptr(20));

to work - you would have to have sth like:

// 'ptr' must return sth (int for example)
// if you want its ret val to be passed as arg to 'function'
// this way you do not have to invoke 'ptr' from within 'function'
int ptr(int);
int function(int, int , int);

Solution 2:

The usual trick is to use a typedef for signature:

 typedef void signature_t (void*, int, const char*);

^{Notice that without the typedef the syntax is like a function declaration. It declares signature_t as a typedef for functions, so you'll always use pointers to signature_t in practice.}

Then you can declare your "high-order" function as

 int function (int, int, signature_t*);

See also this reply.

Solution 3:

The correct syntax for function call is:

function(20,20, &ptr);

If you feel lost, try some tutorials, or this