On Tue, Apr 11, 2006 at 11:44:43PM +0800, Dan Jacobson wrote:
> Package: perl
> Version: 5.8.7-10
> Severity: normal
> File: /usr/bin/a2p
>
> Big fat a2p bug, not mentioned on man page even:
> $ awk 'BEGIN{print 0.6 % 11}'
> 0.6
> $ echo 'BEGIN{print 0.6 % 11}'|a2p|perl
> 0
> $ echo 'BEGIN{print 0.6 % 11}'|a2p|tail -1
> print 0.6 % 11;
> If it is OK that 0.6=0 sometimes then the a2p man page should say why.
>
> And perlop is wrong:Not wrong. Incomplete maybe... > Binary "%" computes the modulus of two numbers. Given integer > operands $a and $b: If $b is positive, then "$a % $b" is $a > minus the largest multiple of $b that is not greater than $a. > As it doesn't mention what happens to non integers!! OK, what about > non-integers?! Say something! That would be because, mathematically, modulus is typically an integer operation. It appears from your experimentation that you have a good idea of what actually happens. My guess is that you could have actually submitted the following patch. ==== //depot/perl/pod/perlop.pod#145 (text) ==== @@ -260,7 +260,9 @@ C<$a> minus the largest multiple of C<$b> that is not greater than C<$a>. If C<$b> is negative, then C<$a % $b> is C<$a> minus the smallest multiple of C<$b> that is not less than C<$a> (i.e. the -result will be less than or equal to zero). +result will be less than or equal to zero). If the operands +C<$a> and C<$b> are floting point values, only the integer portion +of C<$a> and C<$b> will be used in the operation. Note that when C<use integer> is in scope, "%" gives you direct access to the modulus operator as implemented by your C compiler. This operator is not as well defined for negative operands, but it will
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