On 10/11/06, Moon, John <[EMAIL PROTECTED]> wrote:
> perl -e '@a=("frc.apmt","frc_ff.apmt");print join(q{,}, map(&subt($_),
> @a)), "\n";
> sub subt {my ($a) = @_; $a=~s/^(.*)\..*/$1/; print "a=$a\n";
> return $a;}'
> perl -e '@a=("frc.apmt","frc_ff.apmt");print join(q{,},
> map(s/^(.*)\..*/$1/, @a)), "\n"; '
> perl -e '@a=("frc.apmt","frc_ff.apmt");print join(q{,},
> map(s/^(.*)\..*/\1/, @a)), "\n"; '
>
> Can someone explain why the last two examples don't product the same
> output as the first?
>
> Thank you in advance.
> jwm
>>Sure. It's the first par of the sub:
>>
>> my $a = @_;
>>
>>In scalar context, an array returns the number of elements, so your $a
>>gets 1. Even though you only pass your sub 1 item, @_ is still an
>>array. It's a 1-element array, but it's an array. What you want is:
>>
>> my $a = shift @_;
>>
>>Also, for future reference using $a and $b is dangerous and considered
>>bad form; Perl uses them internally to implement sort.
>>
>>HTH
Thank you for looking but first part of sub is:
my ($a) = @_;
and the first test script is returning what I want ... it's the second &
third I'm asking about
jwm
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