CODEWITHSUNDEEP
cbit-manipulationbitwise-operatorsbitwise-not
user11839113
user11839113

Reputation:

~ Unary Operator and Bitwise Tests Give Negative Results

I was studying bitwise operators and they make sense until the unary ~one's complement is used with them. Can anyone explain to me how this works?

For example, these make sense however the rest of the computations aside from these do not:

1&~0 = 1   (~0 is 1 -> 1&1 = 1)
~0^~0 = 0  (~0 is 1 -> 1^1 = 0)
~1^0 = 1   (~1 is 0 -> 0^1 = 1)
~0&1 = 1   (~0 is 1 -> 1&1 = 1)
~0^~1 = 1  (~0 is 1, ~1 is 0 -> 1^0 = 1)
~1^~1 = 0  (~1 is 0 -> 0^0)

The rest of the results produced are negative(or a very large number if unsigned) or contradict the logic I am aware of. For example :

0&~1 = 0   (~1 = 0 therefor 0&0 should equal 0 but they equal 1)
~0&~1 = -2
~1|~0 = -1

etc. Anywhere you can point me to learn about this?

Upvotes: -1

Views: 86

Answers (3)

klutt
klutt

Reputation: 31389

~1 = 0 - No it's not. It's equal to -2. Let's take a eight bit two complement as example. The decimal number 1 has the representation 0000 0001. So ~1 will have 1111 1110 which is the two complement representation of -2.

Upvotes: 0

John Szakmeister
John Szakmeister

Reputation: 47032

They actually do make sense when you expand them out a little more. A few things to be aware of though:

  1. Bitwise AND yields a 1 only when both bits involved are 1. Otherwise, it yields 0. 1 & 1 = 1, 0 & anything = 0.

  2. Bitwise OR yields a 1 when any of the bits in that position are a 1, and 0 only if all bits in that position are 0. 1 | 0 = 1, 1 | 1 = 1, 0 | 0 = 0.

  3. Signed numbers are generally done as two's complement (though a processor does not have to do it that way!). Remember with two's complement, you invert and add 1 to get the magnitude when the highest bit position is a 1.

Assuming a 32-bit integer, you get these results:

 0 & ~1 = 0 & 0xFFFFFFFE = 0
~0 & ~1 = 0xFFFFFFFF & 0xFFFFFFFE = 0xFFFFFFFE (0x00000001 + 1) = -2
~1 | ~0 = 0xFFFFFFFE & 0xFFFFFFFF = 0xFFFFFFFF (0x00000000 + 1) = -1

Upvotes: 1

Blaze
Blaze

Reputation: 16876

0&~1 = 0 (~1 = 0 therefor 0&0 should equal 0 but they equal 1)

~1 equals -2. If you flip all the bits of a Two's Complement number, you multiply it by -1 and subtract 1 from the result. Regardless of that 0 has 0 for all the bits, so the result of & is going to be 0 anyway.

~0&~1 = -2

~0 has all bits set so ~0&~1 is just ~1. Which is -2.

~1|~0 = -1

~0 has all bits set, so the result of the | is ~0 (= -1) no matter what it is OR'd with.

Upvotes: 0

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