Evaluation order in return statement

Question

//Consider the following C program
#include <stdio.h>
int f(int *a,int *b,int c) {
  if(c == 0) return 1;
  else {
    *b = *b-1;
    *a = *a+1;
    c = c - 1;
  return *a + f(a,b,c) + *b;
  }
}
int main() {
  int a = 2,b = 2,c = 2;
  printf("%d\n",f(&a,&b,c));
  return 0;
}

Doubts :

• Is there any undefined behaviour according to C language?
• if NO, then what will be the output ?
• Please explain with proper explanation.

Comments

yes its correct.

---

@air1 what about my command?

any idea?

---

@srestha

int f(int *a,int *b,int c) { 
  /*Some Code*/

  else {
    *b = *b-1; //note that b is int* type variable
    *a = *a+1; //here a is int* type variable
    c = c - 1;

    return *a + f(a,b,c) + *b;

    //so here when we call f(a,b,c) Note: a & b are pointers 
    //of int type so here we are passing pointers to
    //f(int*,int*,c), since C don't support
   //pass-by-reference we simulate it using pointers so this is 
   //call-by-value only but value here is address of an integer
   //in a and b so we are simulating pass by reference with help
   //of pointers.

  } //else
} //f()

Also I think it is sequence point problem since return value of function depends upon order of evaluation.