C Parameter Passing and Pointers

Question

void f(int pv1, int *pv2, int *pv3, int pv4[]){
  int lv = pv1+ *pv2 + *pv3 + pv4[0];
  pv1= 11;
  *pv2= 22;
  *pv3= 33;
  pv4[0]= lv;
  pv4[1]=44;
 }

int main(void){
  int lv1=1, lv2=2;
  int *lv3;
  int lv4[]= {4,5,6};
  lv3= lv4+2;
  f(lv1, &lv2, lv3 , lv4);
  printf("%i,%i,%i\n", lv1, lv2, *lv3);
  printf("%i,%i,%i\n", lv4[0], lv4[1], lv4[2]);
  return 0;
}

Answer : 1 22 33

I don't understand how.

My working: lv1=1, lv2=2, lv3 = lv4+2= 6, lv4[]= {4,5,6}

After going through f(), Lv1=11, lv2=22, lv3=33, lv4[0]=13 and lv4[1]=44

Please Add Some Explanation What You Want To Do So One Can Check The Mistake — Ajay Pandya, Mar 06 '17 at 04:14
Possible duplicate of [How do I modify a pointer that has been passed into a function in C?](http://stackoverflow.com/questions/766893/how-do-i-modify-a-pointer-that-has-been-passed-into-a-function-in-c) — StoryTeller - Unslander Monica, Mar 06 '17 at 10:23

score 1 · Answer 1 · edited Mar 06 '17 at 10:19

1

Nothing is not understandable here.

pv1= 11; will not affect value of lv1, because you are just populating a local variable pv1 in f(), not any pointer. pv1 argument is passed by value, and becomes local to f().
Values of *lv3 & lv2 are updated with 22 & 33. Because, your updating values in the address using pointers.

edited Mar 06 '17 at 10:19

linuxfan says Reinstate Monica

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answered Mar 06 '17 at 04:45

jjm

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C Parameter Passing and Pointers

1 Answers1