Question

I was writing a program to concatenate two arrays in C. I am allocating memory for a third array and using memcpy to copy the bytes from the two arrays to the third. The test output is:

1 2 3 4 5 0 0 0 0 0

Is there anything wrong with this approach?

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

int *array_concat(const void *a, int an,
                   const void *b, int bn)
{
  int *p = malloc(sizeof(int) * (an + bn));
  memcpy(p, a, an*sizeof(int));
  memcpy(p + an*sizeof(int), b, bn*sizeof(int));
  return p;
}

// testing
const int a[] = { 1, 2, 3, 4, 5 };
const int b[] = { 6, 7, 8, 9, 0 };

int main(void)
{
  unsigned int i;

  int *c = array_concat(a, 5, b, 5);

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

  free(c);
  return 0;
}

Answer 1

memcpy(p + an*sizeof(int),...

this second memcpy, you are trying to add 5 * sizeof(int) to an int pointer, p. However, when you add to a pointer, it already knows that it has to deal with sizeof(type), so you don’t have to tell it.

memcpy(p + an,...