How to cast the address of a pointer generically while conforming to the C standard

Question

It is common to assign pointers with allocations using an implicit function-return void * conversion, just like malloc()\'s:

void *malloc(size_t size);
int *         


        

Answer 1

Types int** and void** are not compatible You are casting p, whose real type is int**, to void** and then dereferencing it here:

*((void **) p) = pv;

which will break aliasing rules.

You can either pass a void pointer and then cast it correctly:

void *pi = NULL;
int* ipi = NULL ;
allocate_memory(&pi, sizeof *ipi );
ipi = pi ;

or return a void pointer.

int *pi = allocate_memory(sizeof *pi);

There is an option to use a union:

#include 
#include 
#include 

union Pass
{
    void** p ;
    int** pi ;
} ;

int allocate_memory(union Pass u , size_t s) {
    void *pv;
    if ( ( pv = malloc(s) ) == NULL ) {
        fprintf(stderr, "Error: malloc();");
        return -1;
    }
    printf("pv: %p;\n", pv);
    *(u.p) = pv;

    return 0;
}

int main() 
{
    int* pi = NULL ;
    printf("%p\n" , pi ) ;
    allocate_memory( ( union Pass ){ .pi = &pi } , sizeof( *pi ) ) ;
    printf("%p\n" , pi ) ;

    return 0;
}

As far as I understand it, this example conforms to standard.

Use static asserts to guarantee that the sizes and alignment are the same.

_Static_assert( sizeof( int** ) == sizeof( void** ) , "warning" ) ;
_Static_assert( _Alignof( int** ) == _Alignof( void** ) , "warning" ) ;