What double negation does in C

Question

I had a controversy about what compilers \"think\" about this:

a = 8;
b = !!a;

So, is b == 0x01 ? Is TRUE always

Answer 1

The result of the logical negation operator ! is 0 if the value of its operand compares unequal to 0, 1 if the value of its operand compares equal to 0. The result has type int. The expression !E is equivalent to (0==E). C11 §6.5.3.3 5

b below will typical have the value of 1 (or possible -1, see below).

a = 8;
b = !!a;

So, is b == 0x01?

Yes (* see below)

Is TRUE always 0x01 or it may be 0xFF, 0xFFFF, 0xFFFFFFFF etc..?

In , true is a macro with the integer constant 1. TRUE is not defined by the C standard. Various implementations defines it with the value of 1. It might be have other values. It certainly should be non-zero.

what is better to use?

A)  a>0?1:0 
B)  !!a

Both result in an int with the value of 0 or 1. A reasonable to good compiler would be expected to generate the same code for both (if a in not signed). Use the form that 1) adhere to your groups coding standard or else 2) best conveys the meaning of code at that point. A third option which results in type (bool):

C)  (bool) a

If a is signed , then a>0?1:0 is not equivalent to !!a. a != 0 ? 1 :0 is equivalent of !!a.

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* If b is a 1 bit signed bit field, b = !!8 will have the value of -1.