Question

I fall in some problem.

I need to write some function like memcpy(void*, const void*), which its signature should be:

void arrayCopy(void *dest, int dIndex, const void *src, int sIndex, int len)

I noticed that, in many implementation of memcpy, we cast void* to char*, but I think this is not the case of me, as the arrayCopy function needed to be used on arrays of many types including structs.

So, how can I accomplish this?

EDIT:
the source code might be something like that:

#include <stdio.h>
#include <string.h>

void arrayCopy(void *, int, const void *, int, int, size_t);

int main(void)
{
    int i;
    int dest[10] = {1};
    int src [] = {2, 3, 4, 5, 6};

    arrayCopy(dest, 1, src, 0, 5, sizeof(int));

    for (i=0; i<10; i++) printf("%i\n", dest[i]);

    return 0;
}

void arrayCopy(void *dest, int dIndex, const void *src, int sIndex, int len, size_t size)
{
    char *cdest = (char*) dest;
    const char *csrc = (char*) src;
    int i;

    len *= size;

    if (dest == src)
    {
        printf("Same array\n");
    }else
    {
        cdest += (dIndex * size);
        csrc += (sIndex * size);
        for (i=0; i<len; i++)
            *cdest++ = *csrc++;
    }
}

Thanks.

Answer 1

“char * ” is just a bunch bytes, everything in C is ultimately bytes – you can cast a pointer to any data structure to char* (you will also need to know the size in memory of the structure)