C Programming - Expressions - Discussion

According to me, it is 3.

@All.

The output will be 3.

#include<stdio.h>
int main()
{
int i=3;
i = i+1;
printf("%d\n", i);
return 0;
}

returns 4 because it is assigned at that moment only whereas

#include<stdio.h>
int main()
{
int i=3;
i = i++;
printf("%d\n", i);
return 0;
}

will return 3 because the value assigned is 3 with post increment.

#include<stdio.h>
int main()
{
int i=3;
i = i++;
printf("%d\n", i);
return 0;
}

The output will be only 3 not 4 bcoz of post-increment as i=i++(it means that i will not be increased by 1 while 1 is incremented in memory not in the value of i but if in place of (i=i++) only i++ is given then means that value of i is increment by one )

If code is written like this then output will 4.
#include<stdio.h>
int main()
{
int i=3;
i++;
printf("%d\n", i);
return 0;
}

Thank you.

It is wrong because of i=3;

i=i++;
first i=4;
seound i++=3;
than
i=i++;

So, i++ override the value of i;
The output will be 3;.

The ans is 3 not 4,

Here, i=i++ is divided into temp=i; i=i++; i=temp so the i value remains unchanged.

I tried in the gcc compiler in linux and the output is 3. But here it says 4. How, please explain.

The answer should be 3.

Because the postfix works totally different than prefix.
When postfix operator is used it uses another temporary variable while prefix uses the same variable for increment.

NOTE: the temp most probably will be a register.
So here it is how it goes:
i++ -> temp = 4;
But i = i++ ; which means i++ will evolve after solving the assignment and the assignment goes to temp var.

Hope I was clear.

I just ran it on my computer and it is 3 not 4. It's a post-increment.

@Rishabh.

You're damn right! the answer is 3 in case of such post increment.

i = i++;
is analogous to
temp = i;
i = i + 1;
i = temp;

I think the answer is 3. I run it on both GCC complier or turbo C compiler try it on your own.