C array declaration and assignment?

Question

I\'ve asked a similar question on structs here but I\'m trying to figure out how C handles things like assigning variables and why it isn\'t allowed to assign them to eachot

Answer 1

When saying "int x[10]" is saying, "reserve some room for 10 integers and pass me a pointer to the location". So for the copy to make sense you'd need to operate on the memory pointed by, rather than 'the name of the memory location'.

So for copying here you'd use a for loop or memcpy().

Answer 2

Some messages here say that the name of an array yields the address of its first element. It's not always true:

#include <stdio.h>

int
main(void)
{
  int array[10];

  /*
   * Print the size of the whole array then the size of a pointer to the
   * first element.
   */
  printf("%u %u\n", (unsigned int)sizeof array, (unsigned int)sizeof &array[0]);

  /*
   * You can take the address of array, which gives you a pointer to the whole
   * array. The difference between ``pointer to array'' and ``pointer to the
   * first element of the array'' matters when you're doing pointer arithmetic.
   */
  printf("%p %p\n", (void*)(&array + 1), (void*)(array + 1));

  return 0;
}

Output:

40 4
0xbfbf2ca4 0xbfbf2c80

Answer 3

In order to assign arrays you will have to assign the values inside the array.

ie. x=y is equivalent to

for(int i = 0; i < 10 < ++i)
{
x[i] = y[i];
}

Answer 4

In an attempt to complement Blank's answer, I devised the following program:

localhost:~ david$ cat test.c
#include <stdlib.h>
#include <stdio.h>
int main (int argc, char * argv [])
{
  struct data {
    int c [2];
  } x, y;
  x.c[0] = x.c[1] = 0;
  y.c[0] = y.c[1] = 1;
  printf("x.c %p %i %i\n", x.c, x.c[0], x.c[1]);
  printf("y.c %p %i %i\n", y.c, y.c[0], y.c[1]);
  x = y;
  printf("x.c %p %i %i\n", x.c, x.c[0], x.c[1]);
  printf("y.c %p %i %i\n", y.c, y.c[0], y.c[1]);

  return 0;
}

When executed, the following is output:

x.c 0x7fff5fbff870 0 0
y.c 0x7fff5fbff860 1 1
x.c 0x7fff5fbff870 1 1
y.c 0x7fff5fbff860 1 1

The point is to illustrate how the copy of structures' values occurs.

Answer 5

Simply put, arrays are not assignable. They are a "non-modifiable lvalue". This of course begs the question: why? Please refer to this question for more information:

Why does C++ support memberwise assignment of arrays within structs, but not generally?

Arrays are not pointers. x here does refer to an array, though in many circumstances this "decays" (is implicitly converted) to a pointer to its first element. Likewise, y too is the name of an array, not a pointer.

You can do array assignment within structs:

struct data {
    int arr[10];
};

struct data x = {/* blah */};
struct data y;
y = x;

But you can't do it directly with arrays. Use memcpy.

Answer 6

I've used C compilers where that would compile just fine...and when run the code would make x point to y's array.

You see, in C the name of an array is a pointer that points to the start of the array. In fact, arrays and pointers are essentially interchangable. You can take any pointer and index it like an array.

Back when C was being developed in the early 70's, it was meant for relatively small programs that were barely above assembly language in abstraction. In that environment, it was damn handy to be able to easily go back and forth between array indexing and pointer math. Copying whole arrays of data, on the other hand, was a very expensive thing do do, and hardly something to be encouraged or abstracted away from the user.

Yes, in these modern times it would make way more sense to have the name of the array be shorthand for "the whole array", rather than for "a ponter to the front of the array". However, C wasn't designed in these modern times. If you want a language that was, try Ada. x := y there does exactly what you would expect; it copies one array's contents to the other.