Ever find yourself wrangling in a map when you're trying to change each element of a list to a new value via s/// or tr/// only to recall that substitution or transliteration returns a count, not the changed string? So you end up doing something ugly like: 
@new = map { my $x = $_; $x =~ s/foo/bar/g; $x } @old;

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That's just because you're doing it inside out. Much simpler to use a foreach loop, as shown in this snippet.
s/foo/bar/g for (@new = @old);
- or -
tr/A-Z/N-ZA-M/ for (@new = @old);

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Replies are listed 'Best First'.
(jcwren) RE: make a new list by substituting an old list
by jcwren (Prior) on Sep 10, 2000 at 21:39 UTC
    Is there something wrong with this, then? 
    my @new = map {s/foo/bar/g; $_} @old;

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    --Chris

    e-mail jcwren
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[d/l]
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        See what happens when you let small children play with sharp regexps? <G>

        (I knew I should have asked Ovid about that first...)

        --Chris

        e-mail jcwren
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