Can someone explain how ++ operator works if it is used more than once in the same expression?

Question

#include <iostream>
using namespace std;
int main() {
int i=0;
i=i++ + i++;
cout << i;
int j=0;
j = j++ + j++ + j++;
cout << endl << j ;
return 0;
}
// why do we have 1 and 3 ?
// i expected to see 2 and 3 as an output
https://code.sololearn.com/cV5244tCamDp/?ref=app

Answer 1

i think it working that:
i = 0 + 1++;
then
i is increment and then i = that what is on right side( 1 )

Answer 2

@Manual @Wojciech I like your explanations.
int j=0;
j = j++ + 4*j++ + 3*j++;
I will try to make a quizz from this :) as @Wojciech suggested.

Answer 3

I edited your code and now it shows how it work
https://code.sololearn.com/cMkqIc8y1L6C/?ref=app

Answer 4

This is my take on it.
in the comments of the copied code.
https://code.sololearn.com/cqroPX4TXQWG/?ref=app

Answer 5

And if it was i= i++ + ++i , only then the value would be 2. Check out

Answer 6

@Jeevitha I think in your case
k=0
r = k++ + k++ + k++ + k++
it is calculated like r=0+1+2+3;

Answer 7

And if it was i= i++ + ++i , only then the value would be 2. Check out - you are right :) have fun with c++!

Answer 8

I find it can be tricky.
When I first use them,
I had to play with the values.
I have to try compiling three of them, to recreate your example.

Answer 9

If it is post increment i++, then first value will be assigned and then it will be increased by one..
Here in ur expression i=i++ + i++,
First i will be assign as 0 thn increased by one..now the expression becomes i=0++ + i++
Then the + operator is for addition
Now i has increased it is 1,in ur expression there is again post increament of i i.e. i++ same rule.. First assign thn increase..
So now as i=1 first assign it to the expression thn increase
Now ur expression becomes
i=0++ + 1++(aftr this i increased by 1)
=0 + 1
=1
So finally, ans is i=1
Similarly for j=j++ + j++ + j++
Ans is: j=0++(increase by 1 becomes 1) + 1++ (increase by 1 becomes 2) + 2++(increase by one becomes 3)
J=0+1+2=3
So the output is 1 and 3.
It is very simple to solve these operators problem, if one simply understand two formulas i.e.
In post increament or decreament like x++/x--, FIRST ASSIGN THEN INCREASE/DECREASE
In pre increament/decreament like ++x/--x,
FIRST INCREASE/DECREASE THEN ASSIGN.
THANK YOU...

Answer 10

I added a comment on your code you can check it out

Answer 11

When you say i++ + i++ it's 1+ 0 = 1. Your i value is 0 2nd time coz u r adding immediate i and not incremented i. In simple ignore last i
U can follow my instructions for other examples too. As follows:
J=j++ + j++ + j++; will be 1+2+0 = 3
K=k++ + k++ + k++ + k++; will be 1+2+3+0=6
U can check with 5 increments like this n and ans will b 10. Try out

Answer 12

it will show error