Global and Local Scope Variables ( Why is the 2nd printout 28 here? )

Question

#include <stdio.h>

int i = 3, j = 10;

int crypt(int j)
{
  return (i = i+j);
}

void decrypt(int x, int i)
{
  j += crypt(i);
}

int main(void)
{
  int i = 0;
  i = crypt(5);
  decrypt(i, j);
  printf("|%d %d|", i, j);
  return 0;
}

I'm having trouble figuring out why does it printout |8 28|.

The "8" part, I understand that at

i = crypt(5) -> j is now 5 in this function -> i = i + j -> There's no i therefore it uses the global variable i = 3 -> i = 3 + 5 -> returns i = 8

So the i in the main function becomes 8.

But what about the next printout? Why is it 28 instead of 23?

The way I read it was like this

decrypt(i, j) -> decrypt(8, 10) -> x is now 8 and i is now 10 in this function -> j += crypt(i) -> j += crypt(10) -> j in this function is now 10.

return ( i = i + j ), there's no i in this function so i = 3 + 10... returns 13?

So then j += 13 is 23?

Which part of the step did I mess up? I've been reading local / global scope online and I still don't quite get where did I go wrong... Feels like I'm messing up my value for i somewhere.

PS: I apologize for the poor formatting, not quite sure how else can I put it cleanly.

Answer 1

You write:

return ( i = i + j ), there's no i in this function so i = 3 + 10... returns 13?

No, i is not 3 anymore. It was changed to 8 previously, i.e. here return (i = i+j); due to the first call of crypt

When you write:

So the i in the main function becomes 8.

it's correct but the global i was changed as well.