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Home/ Questions/Q 7177647
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T16:45:28+00:00 2026-05-28T16:45:28+00:00

The following code, I think, describes what I am trying to do. Specifically, I

  • 0

The following code, I think, describes what I am trying to do. Specifically, I wish to cast a function pointer to a generic function type, with the only difference in signature being different pointer types.

Now, I’m aware that there is a requirement for function pointers to be compatible as discussed in this question, but I’m not sure whether having an argument of different pointer type satisfies that compatibility requirement.

The code compiles and runs, but, as expected, gives warnings about the assignment from incompatible pointer type. Is there some way to satisfy the compiler and achieve what I am after?

#include <stdio.h>

int float_function(float *array, int length)
{
    int i;
    for(i=0; i<length; i++){
        printf("%f\n", array[i]);
    }
}

int double_function(double *array, int length)
{
    int i;
    for(i=0; i<length; i++){
        printf("%f\n", array[i]);
    }
}


int main() 
{
    float a[5] = {0.0, 1.0, 2.0, 3.0, 4.0};    
    double b[5] = {0.0, 1.0, 2.0, 3.0, 4.0};

    int (*generic_function)(void*, int) = NULL;

    generic_function = &float_function;
    generic_function(a, 5);

    generic_function = &double_function;
    generic_function(b, 5);

    return 0;
}
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  1. Editorial Team
    Editorial Team
    2026-05-28T16:45:29+00:00Added an answer on May 28, 2026 at 4:45 pm

    The cleanest way is IMHO to perform the cast inside the function. This will force all the function signatures to be the same, and keeps the casts out of the caller’s code. (this is for instance the way that qsort() wants it)

    int double_function(void *p, unsigned size)
    {
        double *array = p    
        unsigned uu;
    
        for(uu=0; uu < size; uu++){
            printf("%f\n", array[uu]);
        }
    return 42;
    }
    
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