siegfr...@heintze.com wrote:
Darn -- I forgot to switch to plain text again. I hope this does not
appear twice -- I apologize if it does!
This works and produces the desired result (I've simplified it a bit):
$default= ((((`grep pat file-name`)[0])=~/[0-9]+/)[0]);
I think you mean either:
$default= ((((`grep pat file-name`)[0])=~/[0-9]+/g)[0]);
Or:
$default= ((`grep pat file-name`)[0])=~/[0-9]/;
Because the match operator /[0-9]+/ will only return true or false in
either list or scalar context
$ perl -le'print "List context: ", "5" =~ /[0-9]+/'
List context: 1
$ perl -le'print "Scalar context: ", scalar( "5" =~ /[0-9]+/ )'
Scalar context: 1
Why does it take so many parentheses?
It doesn't, this will work as well:
$default = ( ( `grep pat file-name` )[ 0 ] =~ /[0-9]+/g )[ 0 ];
Or:
( $default ) = ( `grep pat file-name` )[ 0 ] =~ /[0-9]+/g;
Or even:
$default = ( `grep pat file-name` =~ /[0-9]+/g )[ 0 ];
Or:
( $default ) = `grep pat file-name` =~ /[0-9]+/g;
I don't think it should work, however.
(1) Why cannot I just index the results of the sub-process directly and
say `grep pat file-name`[0]?
Because `grep pat file-name` is not a list (or array).
If Perl is confused I would think I might
need to explicitly convert it like this:
@{`grep pat file-name`}[0]
but that does not work. I think it should.
@{} dereferences an array reference but `grep pat file-name` is not an
array reference.
You could copy the output from `grep pat file-name` to an array or
anonymous array and index that:
[ `grep pat file-name` ]->[ 0 ]
But then you might as well just use a list slice in the first place.
(2) I have the same question about the =~ operator -- it returns an
array too.
The binding operators (=~ and !~) do not "return" anything, they just
bind the expression on their left hand side to the operator on the right
hand side. It is the operator on the right hand side which returns
something. The binding operators can bind to either the match operator
(m//) or the substitution operator (s///) or the transliteration
operator (tr/// or y///).
So why cannot I just type
print @{$a=~/([0-9]+)/}[0] ?
Again, @{} dereferences an array reference and $a=~/([0-9]+)/ is not an
array reference.
Instead I have to type
print (($a=~/([0-9]+/)[0]);
Why are the extra outer parens required?
perldoc -f print
Also be careful not to follow the print keyword with a left
parenthesis unless you want the corresponding right parenthesis to
terminate the arguments to the print--interpose a "+" or put
parentheses around all the arguments.
In other words:
print ($a=~/([0-9]+/)[0];
Is just:
print($a=~/([0-9]+/);
Followed by a syntax error.
John
--
Any intelligent fool can make things bigger and
more complex... It takes a touch of genius -
and a lot of courage to move in the opposite
direction. -- Albert Einstein
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