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I think I understand what youre saying on its own but I must not because I don't see how that contradicts my statement.
Foo returns (q+y) which in this case is (x+y) or (1), and I understand that...

So is q automatically 0 since we never assigned it a value? and is foo(x) the just the value that foo returns if it used x as the value instead of q?

So when it jumps to foo which prints q=0 and...

Okay, help me understand whats going on here.



#include <stdio.h>

int foo (int q) { /* so here we're declaring a function called foo which returns an integer.
(int q) means that our foo...