Pre increment operator

Question

#include<iostream>
using namespace std;
int main()
{
    int i=2;
    cout<<++i<<" "<<++i;
    return 0;
}

Why the output of program is '4 4' not '3 4' ?

score 0 · Accepted Answer · answered Nov 09 '12 at 17:07

0

because both increments occur before the line is output. the actual writing to the screen is delayed until the entire line has been run, but by then the reference for i has already had its value updated.

if you split your cout line into two discrete outputs, things would evaluate as you expect.

answered Nov 09 '12 at 17:07

Frank Thomas

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Does output also depends on the compiler one is using ? – Shubham Jaiswal Nov 09 '12 at 19:51
TBH, I'm not sure, but cout is a stream, and a streams get built and then flushed as a whole, so my guess is that this behavior is inevitable. – Frank Thomas Nov 09 '12 at 21:26

score 0 · Answer 2 · answered Nov 09 '12 at 17:26

You have two side effects on the same variable (the two increments) with no sequence point between them. So they might happen in any order or might even get interleaved -- the behavior is undefined. You appear to be getting interleaved behaviors here -- the expression ++i is increment I then read i. So you're getting increment, increment, read, read.

Pre increment operator

2 Answers2