Question

I’ve the following code.if i give control_word as 6 why if condition evaluates to true and enters inside if block?what exactly is happening here?

#define MACRO1 0x01
#define MACRO2 0x02
#define MACRO4 0x04
#define MACRO3 MACRO1 | MACRO2
#define MACRO7 MACRO4 | MACRO3

int main()
{
    if(control_word == MACRO3 || control_word == MACRO7)
    {
        /*DO SOME OPERATION*/
    }
    else
    {
        /*DO SOMETHING ELSE */
    }

}

Answer 1

The expression

control_word == MACRO3 || control_word == MACRO7

expands to

control_word == MACRO1 | MACRO2 || control_word == MACRO4 | MACRO3

which eventually expands to

control_word == 1 | 2 || control_word == 4 | 1 | 2

looking at the precedence table, you see that operator == has higher precendence than |, which is higher than ||, so the evaluation is:

((control_word == 1) | 2) || ((control_word == 4) | 1 | 2)

which evaluates to

((6 == 1) | 2) || ((6 == 4) | 1 | 2)

which is (6==1 is false, which is treated as 0 in the arithmetic expression — same for 6==4)

((0 | 2) || (0 | 1 | 2)

which is

2 || 3

which is

true

as 2 and 3 are treated as true (non-zero), so you enter the if block, not the else

To preserve the (presumed) intent (and get the result you exprected) you need to protect the expansion of the macros by wrapping their definition in parentheses — note that this is always a good idea to avoid the dissonance between what you think and what is actually happening.