what would be the output and why?

Question

#include <stdio.h>
int x = 20;
int f1() { x = x+10; return x;}
int f2() { x = x-5;  return x;}
int main()
{
  int p = f1() + f2();
  printf ("p = %d", p);
  return 0;
}

Comments

As here x modified 2 times, one time adding with 10 and one time assigning the value, so we cannot say what the output will be. If one variable more than one time modified in one sequence point, then there will be sequence point violation for that program. So, output will be compiler dependent

Answer 1

In computer programming, a global variable is a variable with global scope, meaning that it is visible (hence accessible) throughout the program, unless shadowed.

Answer 2

Undefined behavior as per C standard. 

Why? 

Associativity is only used when there are two or more operators of same precedence.

Associativity of the + operator is left to right, but it doesn’t mean f1() is always called before f2(). The output of following program is in-fact compiler dependent. 

In this case if f1() is called first output is 55. But when f2() is called first output is 40. So output is compiler dependent. If in some program output is independent of which function is called, we may say that will be the output.