When I started this post, it was going to be a revolutionary post, talking about something that was really bothering me. As I have spent more time thinking about this, the answer seems so simple and obvious, still there was a time where I did not get this concept, so I here is a brief post on the topic,

The question is, in straight C, will this work:

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

typedef struct point_t {
  int x,y;
} Point;

int main(int argc, char **argv)
  Point a,b;
  int c[4] = {1, 2, 3, 4};

  b = c;

  printf("b(%p) is now b[x] = %d, b[y] = %dn", &b, b.x, b.y);

  return 0;

In C++ this works no problem. Structs are teated like classes where all members have public scope. Assigning one struct to another simple implements the copy constructor that is created by the C++ language, but what about straight C?

Well, like I said the answer is the obvious one, IT WORKS! The reason this originally confused me was because you CAN’T do this:

   int a[4] = { 1, 2, 3, 4};
   int b[4];

   b = a;

Why not? The compiler knows how big each of these integer arrays are. If you run sizeof(a); and it will give you the same thing as sizeof(b);. Copying an array is as simple as copying over the bytes (bit for bit), using memset() or something similar.

The reason structs work and arrays don’t is simple. Basically we are doing a type check on the object. If the object is a struct of defined type, we know how to do an assignment because we have the size. If the item is an array of integers, we would have to do a size comparison and the compiler doesn’t do that.

Okay, so I hope that explains the obvious, good luck.