Thread: Dispute

what does this code do?

Code:

struct x
{
 int a[5];
 int b[5];
};

struct x y[10];

void rotate(void)
{
 int i;
 struct x temp;

 temp=y[9];

 for (i=9; i; i--)
  y[i]=y[i-1];

 y[0]=temp;
}

i thought the compiler copies the elements from y[i-1] to y[i] (just like memcpy()), my friend means the compiler changes only the references from y[i-1]->y[i]. or, in other case, what does y[i] supplies, the address of the element or the whole struct?

EDIT: I had something, but I believe I misread the question. I'm not going to browse here at 3am anymore...

You are correct. In this case, the compiler makes copies as it rotates the structs through the array.

For your friend to be correct, it would have to look something like this:

Code:

struct x
{
  int a[5];
  int b[5];
};


struct x *y[10];


void rotate(void)
{
 int i;
 struct x *temp;

 for(i=0; i<10; i++)
   y[i] = malloc(sizeof(struct x));

 temp=y[9];

 for (i=9; i; i--)
  y[i]=y[i-1];

 y[0]=temp;
}