Copy an array using pointers

Question

Is this a correct way to copy elements from array origin to array location?

  #include <stdio.h>
   void copy(const int *origin, int *location, int n){
   int i;
   for(i=0;i<n;i++){
    location[i]=origin[i];
    }
  }

Answer 1

  #include <stdio.h>

This function doesn't use anything from <stdio.h>, so don't waste the compiler's time by including it.

   void copy(const int *origin, int *location, int n){

I would put the destination argument first, to be consistent with Standard Library functions such as memcpy. And change n to be a size_t, so it will work with any array.

   int i;
   for(i=0;i<n;i++){

We can reduce the scope of i, by declaring it in the control expression: for (int i = 0; i < n; ++i).

     location[i]=origin[i];

We ought to tell the compiler (with restrict) that the arrays don't overlap.

---

I think it's simpler just to forward to memcpy():

#include <string.h>
void copy(int *restrict dest, int const *restrict origin, size_t count)
{
    memcpy(dest, origin, sizeof *dest * count);
}

But that's so simple that I don't think we want a separate function for it (particularly with such a broad name).