Using the result of s///

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

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Re: Using the result of s/// by hardburn (Abbot) on Apr 30, 2003 at 20:11 UTC
`print $string =~ s/_/ /g;` [download] That line actually parses to something like: `$matches = $string =~ s/_/ /g; print $matches;` [download] That's because `print` is receiving the result returned by the substitution, not the string that is being substituted. And the result returned in a substitution is the number of successful substtutions. ---- I wanted to explore how Perl's closures can be manipulated, and ended up creating an object system by accident. -- Schemer Note: All code is untested, unless otherwise stated	[reply] [d/l] [select]
Re: Using the result of s/// by elusion (Curate) on Apr 30, 2003 at 20:15 UTC
The first example you give prints the result of the substitution (the result of `$string =~ s/_/ /g`). If you look at the documentation for s///, you will see that it returns the number of substitutions made. It's easier to see in code. `$string = 'This_has_underscores'; $result = $string =~ s/_/ /g; print $result;` [download] That's what the first result is equal to. The second example prints the new $string, which is what you want in this case. elusion : http://matt.diephouse.com	[reply] [d/l] [select]
Re: Using the result of s/// by Jenda (Abbot) on Apr 30, 2003 at 20:19 UTC
Because that is how it was designed :-) And rightly so (I say this even though I've just been bitten by this in `@array = map {s/regexp/replace/} split(/,/, $text)`) If the s/// returned the resulting string how would you know whether the regexp matched and whether anything was replaced? `if ($text =~ s/foo/bar/) { ... } else { ... }` [download] Jenda Always code as if the guy who ends up maintaining your code will be a violent psychopath who knows where you live. -- Rick Osborne Edit by castaway: Closed small tag in signature	[reply] [d/l] [select]
Re: Using the result of s/// by Cody Pendant (Prior) on Apr 30, 2003 at 21:01 UTC
Perhaps it would help to think of it like this -- print is print (an expression). Like "print (2+2)" will print 4. So your first example isn't a value or a string, it's an expression that gets evaluated. `print ($string =~ s/_/ /g);` [download] -- “Every bit of code is either naturally related to the problem at hand, or else it's an accidental side effect of the fact that you happened to solve the problem using a digital computer.” M-J D	[reply] [d/l]
Re: Using the result of s/// by Zaxo (Archbishop) on Apr 30, 2003 at 20:21 UTC
In the first case, you print the return value from the substitution. In the second, you print the string the substitution was performed on. The return from s// is a fairly complex matter which depends on context, the /g option, and whether backreference capturing is attempted. See perlop and perlre for details. After Compline, Zaxo	[reply]
Re: Using the result of s/// by artist (Parson) on Apr 30, 2003 at 20:16 UTC
In your code `print $string =~ s/_//g` is acutally `my $matches = $string =~ s/_//g; print $matches;` [download] Substitution provides a return value as number of matches. Thus when proceeded by print statement it provides the 'return value' rather than value of the string. When not proceeded by anything, the return value is simply not captured. artist	[reply] [d/l] [select]