Question

If I create a variable within a new set of curly braces, is that variable popped off the stack on the closing brace, or does it hang out until the end of the function? For example:

void foo() {
   int c[100];
   {
       int d[200];
   }
   //code that takes a while
   return;
}

Will d be taking up memory during the code that takes a while section?

Answer 1

No, braces do not act as a stack frame. In C, braces only denote a naming scope, but nothing gets destroyed nor is anything popped off the stack when control passes out of it.

As a programmer writing code, you can often think of it as if it is a stack frame. The identifiers declared within the braces are only accessible within the braces, so from a programmer’s point of view, it is like they are pushed onto the stack as they are declared and then popped when the scope is exited. However, compilers don’t have to generate code that pushes/pops anything on entry/exit (and generally, they don’t).

Also note that local variables may not use any stack space at all: they could be held in CPU registers or in some other auxiliary storage location, or be optimized away entirely.

So, the d array, in theory, could consume memory for the entire function. However, the compiler may optimize it away, or share its memory with other local variables whose usage lifetimes do not overlap.