((var) &= ~(bit))

Code:

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  ((var) &= ~(bit))

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What exactly is that statement saying?
Is that saying toggle 0 or 1?

Or is that saying "whatever it equals
now it equals 0"?

What is var and what is bit?

The ~ operator flips all the bits. So if bit was set to, let's say, 6 (0110), ~bit would be 9 (1001), if bit was a 4-bit integer. Basically, it's a good way to set one or more bits to 0. In the example where bit is 6, you're turning off 2 bits:

val = 01011011 (hypothetically)
bit = 00000110
~bit = 11111001

01011011
& 11111001
---------------
01011001

To me it's saying:
OMG I must be a macro because I have far more brackets than necessary.

As for a description; It does the opposite of:

Code:

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((var) |= (bit))

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:devil:

thats why there so many brackets.
[/code]utils.h:186:#define REMOVE_BIT(var,bit) ((var) &= ~(bit))[/code]

ch->char_specials.saved.affected_by) ( 1 << 8)

but if
ch->char_specials.saved.affected_by doesnt equal (1<<8)

then functions that depend on 1<<8 wont be in affect.
lets say after

~(00001000)

what would that be?

Quote:

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Originally Posted by errigour

~(00001000)

that what would it be?

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hmm...

Quote:

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Originally Posted by itsme86

The ~ operator flips all the bits.

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i meant to ask what that number would be after the ~ mark.

Quote:

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Originally Posted by errigour

i meant to ask what that number would be after the ~ mark.

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You know how to flip bits right? Turn a zero into a one or a one into a zero.

So if only a single bit was a one originally, then only a single bit will be a ____ afterwards. Therefore the value will be ________.

Quote:

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Originally Posted by iMalc

You know how to flip bits right?

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While some prefer a fork, I've always found a spatchula to be most effective...

Quote:

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Originally Posted by errigour

i meant to ask what that number would be after the ~ mark.

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It depends on the size of the value you're flipping. If it's 32 bits in size, then you'd get more 0's turned into 1's, giving you a larger "number".

If you were dealing with 4-bit numbers then ~0x2 would be 1101 (13, 0xD), but if you're dealing with 8-bit numbers then ~0x2 would be 11111101 (253, 0xFD).

Is that what you're asking?

Quote:

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Originally Posted by errigour

i meant to ask what that number would be after the ~ mark.

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For even more fun, try this:

Code:

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printf("%d",(var+(~var + 1));

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;)

whould it flip this intiger

00000100 00000100
to this
00100000 00100000

Quote:

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Originally Posted by errigour

whould it flip this intiger

00000100 00000100
to this
00100000 00100000

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Why in god's name would it? It just turns all the 0 bits into 1s and all the 1 bits into 0s. It would turn 00000100 00000100 into 11111011 11111011.