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# 8  
Old 03-05-2008
So, if I wanted to perform (x & 1) on nawk, I use:

x-(int(x/1)*1) [nawk]
Last edited by DeltaX; 03-05-2008 at 06:37 PM..
# 9  
Old 03-05-2008
Quote:
Originally Posted by Perderabo
And'ing a number with FFFF is the same as finding the remainder when dividing by 65535. This operation is called modulo and could be implemented as:
$ echo "65536
> 65537
> 65538" | awk 'function mod(v,m) { return v-(int(v/m)*m) }
> { print $1, mod($1,65535) }'
65536 1
65537 2
65538 3
$
Cant I use v % m instead of v-(int(v/m)*m) ?
# 10  
Old 03-06-2008
Yep! And it should be 65536. FFFF aka 65535 is the max remainder.
# 11  
Old 03-06-2008
Quote:
Originally Posted by Perderabo
Yep! And it should be 65536. FFFF aka 65535 is the max remainder.
Thanks very much, Perderabo. I just figured that out this afternoon. Its the the operand of the AND plus one.
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