C Programming - Expressions - Discussion

I think the correct answer is 3.

@All.

Let's go through the code step by step to understand why:

int i = 3;: Initializes the integer variable i with the value 3.
i = i++;: This statement is a bit tricky and involves undefined behaviour in C. It is an example of using the post-increment operator (i++) in an assignment.
Here's what happens in detail:

The value of i is read, which is 3.
The post-increment operator i++ is used, which increments the value of 'i' but returns the old value (3 in this case).
The result of the post-increment operation (3) is assigned back to i. So, 'i' is now set to 3.

Due to the undefined behaviour, the C standard does not specify the order of evaluation for the two side effects (increment and assignment) on the same variable within the same sequence point.

Therefore, the output of printf("%d\n", i); will be 3.

It's a post-increment. So the right answer is 3.

3 is the Right Answer.

Correct answer is 4.
because i=3
i=i++ // this is post increment so i=3 but internally i is increment by 1.
so i=4.

As per my knowledge the coding part is;

#include<stdio.h>
int main()
{
int a=(1,2,3);
b=(++a,++a,++a);
c=(b++,b++,c++);
printf("%d %d %d \n", a,b,c);
return 0;
}

Yes, agree the answer is 3.

#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.

The ans is 3 not 4,

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

The answer is 3 because post-increment.