🛑Urgent🛑: Can anyone explain me that how compiler execute this? [Read Description] 🥺🥺 Please help! 😔

Question

#include<stdio.h>
int main()
{
int i = 1;
int j = i++ + (i = i + 5) + i++ + i;
printf("%d", j);
return 0;
}
Output: 23
Why this output?

Answer 1

It produce a error.becuse you can not intilize variable more than one time. And output 23
Here is answer remember i=1
j=i++ +(i=i+5)+ i++ +i;
2+(2+5)+7+7;
2+7+7+7;
21+2;
23.

Answer 2

Angelo
Consider i=1
j = (i=i+2) + i++ + i;
What will be the j's value?
Here,
3 + 3 + 4
Then after this,
Why compiler updates the first term again?
4 + 3 + 4 = 11
Output is 11.

Answer 3

I++=1;// i become 2 (increment)
i=2+5; //now I assign 7
I++=7;// now I assign 8
I=8; // final output by adding all that is 23;

Answer 4

OMG
Undefined behavior, Undefined behavior everywhere!
It is doing:
i=1
+1
i=2
(i=2+5)—>+7
i=7
+7
i=8
+8

Answer 5

Angelo
#include<stdio.h>
int main()
{
int i = 1;
int a = i++ + (i = i + 5) + (i = i + 5);
printf("%d", a);
return 0;
}
Output: 20
Here,
Why second term was not updated again after third term?
I was expecting 25.
I am just confused.

Answer 6

Angelo
#include<stdio.h>
int main()
{
int i = 1;
int a = (i = i + 2) + i++ + i;
printf("%d", a);
return 0;
}
Why output 11?

Answer 7

It's normal that you are confused, we don't really know what the compiler is doing...
That's the meaning of undefined behavior

Answer 8

Ok, I think it is doing some think like:
extract all the i++
substitute them with i-n
So:
i++ + (i=i+5) + i++ + i
becomes
i++; i++ —> i=3
(i-1) + (i=i+5)+(i-2)+i
—>23
(i=i+2) + i++ + i
Becomes
i++ —> i=2
(i=i+2)+(i-1)+i
—>11
i++ +(i=i+5)+(i=i+5)
Becomes
i++ —> i=2
(i-1)+(i=i+5)+(i=i+5)
—>20
As you see I can only make hypothesis that are valid as long as you find an exemple that breaks it (just like with science)

Answer 9

NITIN
According to your logic,
#include<stdio.h>
int main()
{
int i = 1;
int a = (i = i + 2) + i++ + i;
printf("%d", a);
return 0;
}
This program's outout should be 10.
But, why output is 11?

Answer 10

Martin Taylor I'm not sure, C/C++ standard should have a precise operator precedence table, so the result should be unic.
Mahima Rajvir Singh this is the operations' order:
First of all, i = 1
j = i++ so (j = 1) then (i = 2)
j += (i = i + 5) so (i = 7) then (j = 8)
j += i++ so (j = 15) then (i = 8)
j += i so (j = 23) and still (i = 8)
Hope I've been clear.

Answer 11

Sébastien Ranganayaguy , OrHy3 & Dinesh Rathod 👇🏻
According to your logic,
#include<stdio.h>
int main()
{
int i = 1;
int a = (i = i + 2) + i++ + i;
printf("%d", a);
return 0;
}
This program's outout should be 10.
But, why output is 11?

Answer 12

OrHy3
So why it isn't the same in this program?
This is what you have explained.
First of all, i = 1
j = i++ so (j = 1) then (i = 2)
j += (i = i + 5) so (i = 7) then (j = 8)
j += i++ so (j = 15) then (i = 8)
j += i so (j = 23) and still (i = 8)
Thanks for listening my doubts! 😊

Answer 13

Btw, I see you're experimenting a lot with the language
That's great, keep it up!

Answer 14

Output 20 can be:
i=1
+1
i=2
+7
i=7
+12
i=12

Answer 15

J=1+7+7+8;

Answer 16

I try that code in java and the output is 10 I don't now about 11;

Answer 17

https://code.sololearn.com/ckw1aRd6P1ak/?ref=app

Answer 18

And I am not sure but it's because of Operators Precedence
I think you need to read that out then you will understand

Answer 19

NITIN you are a genius!!!
We are trying to understand how C++ works and you talk about Java!
GENIUS!

Answer 20

I think : when u use i++ the value of i will change for the next time u call i.
i++ + (i = i + 5) + i++ + i;
So at the first i++, i = 1
But at i = i + 5, i = 2 for the calcul.
At the seond i++, i = 7,
Then the last part + i, i = 8.
Finally u have : j = 1 + (2 + 5) + 7 + 8 = 23.