Local stacks, curly braces

[bennyandthejets]

If you place curly braces within a function, does that create a new sub-stack? Eg:

int main()
{
    int x;
    {
        int y;
    }
    return 0;
}

Is y in a sub-stack of the one that x is in? That would seem logical, given the behavior of the variables. Ie, x is available everywhere, but y is only available within the nested braces.

Am I on the right track here?

[master5001]

As far as I know scoping rules do not suggest a sub-stack. Just because in your example y can only be used in the inner most set of {}'s does not mean that it is in a different stack.

[thefroggy]

While there is no "sub-stack", scoping can have an effect on the stack. After a variable goes out of scope, the compiler is allowed to use that variable's stack space for variables that are declared later on.

For example:

#include <iostream>

using namespace std;

int main()
{

  {
    int y = 10;
  }

  int x;

  cout << x << endl;

  return 0;
}

My output from this program is 10 even though I never initialize x. The compiler is using y's stack space for x because y will never be in scope again.

Note that you shouldn't depend on this behavior. It may vary depending on which compiler you use and the level of optimization.