Copying arrays of structs in C

Question

It's been a long since I don't use C language, and this is driving me crazy. I have an array of structs, and I need to create a function which will copy one array to another (I need an exact copy), but I don't know how to define the function call. I guess I need to use pointers, but when I try it gives me an error.

struct group{
    int weight;
    int x_pos;
    int y_pos;
    int width;
    int height;
};

struct group a[4];
struct group b[4];

copySolution(&a, &b);

That last declaration send me an error. As I said, it's been a long since programming in C, so I'm a bit lost right now :(

Answer 1

That should do it:

memcpy(&b, &a, sizeof(a));

EDIT : By the way: It will save a copy of a in b .

Answer 2

As Johannes Weiß says, memcpy() is a good solution.

I just want to point out that you can copy structs like normal types:

for (i=0; i<4; i++) {
    b[i] = a[i]; /* copy the whole struct a[i] to b[i] */
}

Answer 3

This has a feel of poorly masqueraded homework assignment... Anyway, given the predetermined call format for copySolution in the original post, the proper definition of copySolution would look as follows

void copySolution(struct group (*a)[4], struct group (*b)[4])
{
  /* whatever */
}

Now, inside the copySolution you can copy the arrays in any way you prefer. Either use a cycle

void copySolution(struct group (*a)[4], struct group (*b)[4])
{
  const struct group *pb, *pbe;
  struct group *pa;

  for (pa = *a, pb = *b, pbe = pb + sizeof *b / sizeof **b; 
       pb != pbe; 
       ++pa, ++pb)
    *pa = *pb;
}

or use memcpy as suggested above

void copySolution(struct group (*a)[4], struct group (*b)[4])
{
  memcpy(b, a, sizeof *b);
}

Of course, you have to decide first which direction you want your arrays to be copied in. You provided no information, so everyone just jumped to some conclusion.

Answer 4

In my case previous solutions did not work properly! For example, @Johannes Weiß's solution did not copy "enough" data (it copied about half of the first element).
So in case, somebody needs a solution, that will give you correct results, here it is:

int i, n = 50;
struct YourStruct *a, *b;

a = calloc(n, sizeof(*a));
b = malloc(n * sizeof(*b));
for (i = 0; i < n; ++i) { 
    // filling a
}

memcpy(b, a, n * sizeof(*a)); // <----- see memcpy here

if (a != NULL) free(a);
a = calloc(n*2, sizeof(*a));

memcpy(a, b, n * sizeof(*b)); // <------ see memcpy here again

Some notes, I used calloc for a, because in the '// filling a' part I was doing operations that required initialized data.

Answer 5

The compiler has no information about the size of the array after passing them as pointer into a function. So you often need a third parameter: The size of the arrays to copy.

A solution (without any error checking) could be:

void copySolution(struct group* a, struct group* b, size_t size) {
    memcpy(a, b, size * sizeof(*a));
} 

Answer 6

The easiest way is probably

 b=a

although a solution with memcpy() will also work.