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开发者_Python百科 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.
"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)
The function must have an element-size info, eg:
void *arrayCopy(void *dest, size_t di,const void *src, size_t si, size_t num, size_t esize)
{
char *cdest = (char*) dest;
const char *csrc = (char*) src;
return memcpy( &cdest[esize*di], &csrc[esize*si], esize*num );
}
...
arrayCopy(dest, 1, src, 0, 5, sizeof*src);
You cannot work with objects of type void
. The Standard doesn't allow that. So you need to cast the void away, and the best type to use is unsigned char
. There's a guarantee by the Standard that unsigned char
can access all bits of any other type representable in your system.
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