X = 5; x=x++ *2+3* --x;

Answer 1

shivam sandhan

Here x++ is post increment which first assign value then increment counter.

--x is post decrement which first decrement value by 1 then assign.

So here

x = x++ * 2 + 3 * --x;

Here x++ will assign value then x will be increased by 1 so x will be 6

x = 5 * 2 + 3 * --x;

As x was 6 so --x will be 5

So now

x = 5 * 2 + 3 * 5 = 25

Answer 2

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Answer 3

That's undefined behavior, the result depends on the compiler