Why the output is a=3 b=5 and not a=3 b=4 for the code:

Question

#include <stdio.h>
int main(){
int a=2, b=3;
b=a++ + b--;
printf("a=%d\tb=%d,a,b);

Answer 1

There's a sequence point in the expression which leads to an undefined behaviour .

Answer 2

Initially a=2 and b=3
And when b=a++ + b-- //these are post increment and post decrement operations
then value becomes b=2 + 3 //first value will assigned then increment(or)decrement will perform. So,
b=5, a=3

Answer 3

Tom Tanner Nope b value is 5

Answer 4

It's one of the classic undefined behavior, but if you understand how STACK works, you know what's when. When you do b = a++ + b--;, a won't be 3 until it meets a again (as in printf), the same goes for b, b won't be 2 unless it meets b again, so the value used is b = 2 + 3;, BUT b-- will do nothing since you are currently doing an assignment to b, therefore the old b (b--) is not visually seen anymore. So b = 5, a still incremented because there is no "replacement" for a, and it meets a again in printf, so it becomes 3.

If you want b to be 4, try b = a++ + --b;
Remember, this is undefined behavior, and different compilers will react differently.

Answer 5

Then b-- will make b's value 4 not 5