why the result is 43

Question

#include <iostream>
using namespace std;

int main() {
int i=2;
i+=4;
i--;
cout <<i--<<--i;
return 0;
}

//Like it if u love it

Answer 1

this is just undefined behaviour, no need to worry, try in another compiler and u will get a different surprise

Answer 2

Guess what. If you do this:

int i=2;
i+=4;
i--;
cout << i-- << i--;

Output would be 45.
So yes, smells fishy.

Answer 3

"The order in which your x++ and ++x statements are evaluated is undefined, so is the effect of your code."
Source: http://stackoverflow.com/a/33445861

Answer 4

,. cout<<i--<<--i as I understand it is two statements not one.. therefore it isnt undefined behaviour at all, it is behaving exactly as expected

it is shorthand for
cout << i--;
cout << --i;

Answer 5

2+4=6 right... 6-1=5.. 5-1=4... 4-1=3..
you couted 4 and 3 to the same line with no spaces this displays them as 43 ..

you could just ask about prefix and postfix operators or just google it...

Answer 6

Using more than one increment or decrement operator with cout in single line leads to undefined behavior.

In you case, output is unexpected because order of evaluation of i-- and --i in single line is undefined.

Answer 7

@jay your answer is not correct. If it wasn't undefined beahvior, it should have printed 53 not 43. Maybe you also need to google postfix and prefix operators :p

Answer 8

@jay it is undefined behavior because as i explained in my answer, order of evaluation of decrement or increment operators used like this in single line is undefined.

If it wasn't undefined, output should be 53 not 43.

Answer 9

output is 53

Answer 10

this is undefined behaviour. but for most compilers this undefined behaves like this. evaluate from right to left. also remember that prefix is lvalue. so it's value can be modified later too