Online test series

Question

#include<stdio.h>
int f(int x,int *py,int **ppz){
  int y,z;
  **ppz+=1;
  z=**ppz;
  *py+=2;
  y=*py;
  x+=3;
  return(x+y+z);
}

void main(){
  int c,*b,**a;
  c=4;
  b=&c;
  a=&b;
  printf("%d",f(&c,b,a));
  return 0;
}

 

Here why doesn't the value of x increment even though the pointers are changed by value?

Why is the answer 19 and not 22?

Comments

In function call you use &c I think that is wrong if you use only 'c' in that place then ans will be 19

(5+7+7)=19

---

Shouldn't it give a compile time error? 

f ( &c, b, a ) mismatches with function header int f ( int x, int *py, int **ppz )

Also there is return in void main function.

Answer 1

the answer is 19 because we are passing the address of c  but there is no pointer in the formal argument it is just declared as a variable  x.so in the called function code works as

ppz and py are the pointer which  will be pointing to c only.

z=5

y=7

x=7

so sum will be 19

Comments

How do you know that x would be 7 only it could be anything as long as you are passing the address ?

---

in the formal arguments we are passing x which will take value as 4 so x+3 means 4+3=7

---

We are passing address of c ? isn`t it?

---

Even if we pass value of C then also the answer will be 19. I think the question has a typo.

  printf("%d",f(c,b,a));  should be used instead of printf("%d",f(&c,b,a));

---

Yes the line should be printf("%d",f(c,b,a));

Answer 2

In function call you use &c I think that is wrong if you use only 'c' in that place then ans will be 19

(5+7+7)=19

Here &c calls the address of the c variable and we want to use it as formal argument to copy value of c to x only so use f(c,b,a)