C Programming - Expressions - Discussion

Good explanation, thanks all.

@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.

I think the correct answer is 3.

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

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;
}

3 is the Right Answer.

Yes, agree the answer is 3.

Answer will be 3.

i=i++
i=3++
I take the value 3 there won't be an increment.
Priority will be given to i=3 not to i++.

Post increment -- after assigning the value to the variable the value is incremented. i.e 3.