+ 2

Doubt in increment operator

If x is already incremented in the if statement, then why does cout<<x give result 1 in the output (ideally the value of x in the memory has become 3 so 33 should be the answer) Can someone tell me where I am going wrong? Check out this code: https://code.sololearn.com/cF1V1QRl4qmU/?ref=app

6th Jun 2017, 6:02 PM
Naman Gogia
Naman Gogia - avatar
2 odpowiedzi
+ 3
This is a nice example of variable shadowing. https://en.wikipedia.org/wiki/Variable_shadowing By doing int x = 2 inside the if statement you are 'hiding' the x previously declared outside the if statement. If we replace the second x by another name, it's easier to understand: if (x>0) { int notX=2; cout<<++notX; } This is basically what you are doing here. Fix: Replace int x = 2; with x = 2;
6th Jun 2017, 6:26 PM
Dennis
Dennis - avatar
+ 2
They are two different variables. The first x's value stays 1 and is never touched.
6th Jun 2017, 6:22 PM
merkrafter