Bryan R Harris wrote:
>
>>>returns "true" or "false" (1 or '') and in list context it returns the
>>>contents of any capturing parentheses in the pattern.
>>>
>>>The expression:
>>>
>>>( $ptypeline =~ /movable.+(sine|geo|radial|ortho)/i
>>>)[ 0 ]
>>>
>>>is a list slice so the regular expression is in list context but the slice is
>>>a single value so the expression is a scalar.
>>>
>>>The || operator will only work with scalar values, not with lists, so this
>>>works because the list has been converted to a scalar with the list slice.
>>>
>>>John
>>********************************************
>>the list context represents everything between the / /
>>and the slice context represents [ 0 ] which is
>>assigned as a scalar to $ptypeline.
>>
>>Correct?
>
>
> Any time you surround something with parenthesis () it is considered "list
> context", i.e.
>
> Scalar context: $a = $ptypeline =~ /movable.+(sine|geo|radial|ortho)/i;
>
> In scalar context, perl is trying to assign a scalar to $a. In scalar
> context that expression returns a 1 or 0 depending on whether it was able to
> find that regular expression inside of $ptypeline. (Or if I had a /gi at
> the end it would return the number of matches it found).
>
> List context: @a = ($ptypeline =~ /movable.+(sine|geo|radial|ortho)/i);
>
> This is list context, meaning that perl is trying to get a list out of that
> expression. In list context, that expression returns whatever items it
> found in sets of parenthesis -- in this case, if ptypeline had "ortho", @a
> would be ("ortho").
No. It is list context because "@a =" forces list context. In other words:
@a = ( $ptypeline =~ /movable.+(sine|geo|radial|ortho)/i );
and:
@a = $ptypeline =~ /movable.+(sine|geo|radial|ortho)/i;
are both in list context, the parentheses are superfluous. However in:
$a = $ptypeline =~ /movable.+(sine|geo|radial|ortho)/i;
the expression is in scalar context because "$a =" forces scalar context.
Even if you add parentheses:
$a = ( $ptypeline =~ /movable.+(sine|geo|radial|ortho)/i );
it is still in scalar context, while:
( $a ) = $ptypeline =~ /movable.+(sine|geo|radial|ortho)/i;
is in list context because $a is now part of a list.
John
--
Perl isn't a toolbox, but a small machine shop where you can special-order
certain sorts of tools at low cost and in short order. -- Larry Wall
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