Why function with no return value modify string outside of its stack

Question

I'm a new in learning C and I do not completely understand how 'void' function can modify some variables. For example

void copyString (char  to[], char  from[])
{
int  i;
for ( i = 0;  from[i] != '\0';  ++i )
to[i] = from[i];
to[i] = '\0';
}

Why can I use modified version of 'to' string? Shouldn't it be modified only in a copyString's stack and not for the whole programm or I misunderstand something? I understand that 'to' is a formal parameter, but I thought that it shoul change the value only inside the function because it is local to that function. Please explain where the problem in my logic?

Answer 1

What seemingly looks like pass-by-value turns into pass-by-reference due to array decay. Array decay is the loss of type and dimension of the array when passed to a function as a value. So instead of passing the array, pointer to the first address of the array is passed.

So what looks like this:

void copyString (char  to[], char  from[])

The Compiler actually looks at it like this:

void copyString (char  *to, char  *from)

You are modifying the values pointed by the pointer rather than a copy in the stack.

More information on array decay can found here.