Definitely a WTF

kdean

No matter what the integer size, this would work a whole lot better:

unsigned long reverse(unsigned long theWord)
{
 unsigned long result = 0;

 int i; 

 for (i = 0; i < sizeof(theWord) * 8; ++i) {
 result <<= 1;
 result |= theWord & 0x01;

 theWord >>= 1;
 }

 return result;
}

kdean

Whoops, didn't mean to create a new thread. Too new to the forums...

Or you could replace the for-loop with a "while (theWord > 0)"

Iago

Aha, but you're assuming a char is 8 bits! :p

You want to use CHAR_BITS, which should be in the standard header limits.h. (You know... just in case someone decides to run your code on a PDP-8...)

maratcolumn1

Oh, not again. Don't know what you're answering to, but you need 5*N =
O(N) ALU operations in order to reverse order of 2^N bits, bytes or
whatever. You used O(2^N). Even with loop unrolling it's 256 operations
instead of 30. Go do your homework again.

BTW "while (theWord > 0)" is incorrect.

kdean

No, you can't replace the for-loop with a "while (theWord > 0)". If theWord = 1, the reversal would stop after one iteration.

kdean

Damn, you're right about the byte size. Replace 8 with CHAR_BITS.

craiga

@kdean said:

Damn, you're right about the byte size. Replace 8 with CHAR_BITS.

That's an even better WTF! The size of long depends on the architecture you are using, whereas the size of char depends on the character set of the application. Imagine a unicode (16 bit) aware application on a 32 bit architecture and you'll see what I mean ...