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        riddles >> cs >> REVERSE  bits in 8 byte 
        
(Message started by: puzzlecracker on Dec 17^th, 2004, 11:33pm)

Title: REVERSE bits in 8 byte
Post by puzzlecracker on Dec 17^th, 2004, 11:33pm

Using C, but without loops or lookup tables, write a routine which reverses the bits in an 8-bit byte

Title: Re: REVERSE bits in 8 byte
Post by John_Gaughan on Dec 18^th, 2004, 10:48am

This should do the trick. Can anyone make it shorter?
::
[hide]unsigned char reverse (unsigned char n)
{
return ((n & 0x01) << 7)
| ((n & 0x02) << 5)
| ((n & 0x04) << 3)
| ((n & 0x08) << 1)
| ((n & 0x10) >> 1)
| ((n & 0x20) >> 3)
| ((n & 0x40) >> 5)
| ((n & 0x80) >> 7);
}
[/hide]::

Title: Re: REVERSE bits in 8 byte
Post by towr on Dec 18^th, 2004, 2:07pm

something like
::[hide]
n = n&11110000 >>4 | n&00001111 << 4;
n = n&11001100 >>2 | n&00110011 << 2;
return n&10101010 >>1 | n&01010101 << 1;

I didn't actually bothered to check whether this does what I think it ought to do, but if it doesn't it should give you the idea of what does..
[/hide]::

Title: Re: REVERSE bits in 8 byte
Post by puzzlecracker on Dec 19^th, 2004, 1:39pm

can both of you, or anyone else pretty much, explain that... thx

I thought to actually flipping the bits as[hide]
n=n^255, but only flips them.... [/hide]

I was wrong... please explain the logic.. thx

Title: Re: REVERSE bits in 8 byte
Post by puzzlecracker on Dec 19^th, 2004, 3:23pm

never mind (i am an idiot) -- towr - you're a genous of modern creation!