Output Tracing of Function Parameter Passing

Question

Output Tracing of C++ program.

How the parameter value is passing in function argument I cann't understand?

#include <bits/stdc++.h>

void call(int,int,int,int);
int main(){
  int a=10;
  call(a,a++,++a,a);
  return 0;
}
void call(int x,int y,int z,int p){
  printf("%d %d %d %d",x,y,z,p);
}

Output:12 11 12 12 But cann't understand the logic behind this.

score 0 · Answer 1 · answered Oct 31 '19 at 03:57

0

++val will increase the value before doing the comparison.
val++ will increase the value after the comparison succeeds.
Code

Arg1 = a
Arg2 = a
a += 1
a += 1
Arg3 = a
Arg4 = a

answered Oct 31 '19 at 03:57

Pickle Rick

808
3
6

I also know that but how the fourth parameter that is passed as 10 becomes 12? – ashik Oct 31 '19 at 04:02
Well it's initialized as 10 and looking through my outline above you can see why. It's incremented twice (once via ++n and once via n++). – Pickle Rick Oct 31 '19 at 04:04
How about the others? – ashik Oct 31 '19 at 04:08
Well output 1 should be 10, all the others are as expected. – Pickle Rick Oct 31 '19 at 04:10

Output Tracing of Function Parameter Passing

1 Answers1