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Re: Using the result of s///

by elusion (Curate)
on Apr 30, 2003 at 20:15 UTC ( #254449=note: print w/ replies, xml ) Need Help??

in reply to Using the result of s///

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;
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 :

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