$ perl -e '$b="H";( $a = "hello" ) =~ s/^\l$b/Y/; print $a'
Yello

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see perlre: \l   lowercase next char

So, after interpolation, "\l$b" gives "\lH" ie "h" and you obtain s/h/Y/

In reply to Re: How to use \l and \u in regex mathing patterns? by brx
in thread How to use \l and \u in regex mathing patterns? by pat_mc

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