behavior of m// operator

[Nathan]

According to the documentation this program should print nothing since
m// uses the last successful pattern match from m//, split, etc. However
on my linux box it prints 1:

#!/usr/bin/perl -w

$a = "a";
$b = "b";
split /a/,$a;
print $b =~ m//;

Is the documentation wrong?

-Nathan

 

[Zak B. Elep]

According to the documentation this program should print nothing since
m// uses the last successful pattern match from m//, split, etc. However
on my linux box it prints 1:

#!/usr/bin/perl -w

$a = "a";
$b = "b";
split /a/,$a;
print $b =~ m//;

Is the documentation wrong?

No, it is correct. `print' emits 1 because the succeeding match
operation ($b =~ m//) succeeded (m// essentially matches a `null'
string, which $b has:

,----[ perl -de 0 ]
| Loading DB routines from perl5db.pl version 1.28
| Editor support available.
|
| Enter h or `h h' for help, or `man perldebug' for more help.
|
| main:-e:1): 0
| DB<1> ($a, $b) = ( qw(a b) )
|
| DB<2> x split /a/ => $a
| empty array
| DB<3> print $b =~ m//
| 1
| DB<4> x $b =~ m//
| 0 1
| DB<5>
`----

Because you used `=~', m// tries to match from $b, not from $a (really
$_) as you would expect.

 

[xhoster]

According to the documentation this program should print nothing since
m// uses the last successful pattern match from m//, split, etc.

My docs don't explicitly say whether the regex used in a split counts
as previously "matched" or not in this context. Experimentally, it looks
to me like it does not.

Xho

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