Programming closed

Question

#include<stdio.h>
void main(){
int i=11;
printf("%d \n", i++|++i);
printf("%d",i);
}

Output is: 15
13

I understand how 13 came but I don't get how 15 came. And why there is no sequence point problem in this?

Answer 1

bro take this line 
printf("%d \n", i++|++i); this is nothing but 

11 = 1011
12 = 1100

11 || 12  = (1011 | 1100) = BitwiseOR(1011,1100) = (1111)₂ = (15)₁₀

Comments

What is the order of evaluation: ++i then i++ "or" i++ then ++i? According to precedence table unary operators are evaluated right to left. Does the first one evaluated in any of them save the changes in the original 'i' variable or not?

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Reference

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Thank you. Could you please check out this code:

jdoodle.com/a/qvv

I do not understand how its output is 13.

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why do u think, there will be no sequence point problem?

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Because we are only evaluating one expression in printf. For example,

printf("%d %d",++i|i++,++i);

this would be a sequence point problem because it involves two expressions to be evaluated on a single variable without using valid sequence point(s).

Am I wrong or right?!

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i think it's fair to say the output is compiler implementation dependent, do you agree @srestha ?

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@Akhilesh Singla,please go to my comment section in the original question and check please. I hope it helps u.