CODEWITHSUNDEEP
c
The_Senate
The_Senate

Reputation: 153

Why does the logical statement !0x00 return 0x01 in C?

I'm reading Computer Systems a programmers perspective, and I'm getting into logical operators, which are similar to bitwise operators, but with a few differences.

What I CANNOT figure out is that when you have a logical operand !0x00 returns 0x01 rather than 0x11?

! is NOT, right? So NOT 0(false) should be 1(true) and another NOT 0(false) should also be 1(true) as well, right?

I look at the bitwise operator example : ~00, naturally that would return 11, but C's logical operators seem to work with vast differences.

Why does this happen?

What I have tried already: Reading a little further to find the answer I seek, it doesn't seem to be here.

What I think the problem is: Might have something to do with how Hexadecimals work? But, Hexadecimals can still have 0x11. . . .

Upvotes: 0

Views: 1611

Answers (3)

Israel Figueirido
Israel Figueirido

Reputation: 343

The expression !n is equivalent to (n==0). It returns 1 if true, 0 if false.

Upvotes: 2

Petr Skocik
Petr Skocik

Reputation: 60097

Neither !0x00 nor ~0x00 will give you 0x11.

!n is the same as n==0 and evaluates to 1 or 0.

~n negates the bits of the binary representation, not the digits of the literal you chose to enter the number.

Upvotes: 0

Lundin
Lundin

Reputation: 214300

Because that's how the language is defined. ! is the logical NOT operator and boolean logic in C works on 1 and 0, representing true and false.

C17 6.5.3.3:

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.

You can think of it as if returning bool, though it actually returns int for backwards-compatibility reasons. Unlike C++ where it does return bool. The same goes for relational and equality operators.

Upvotes: 6

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