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replace s///r ?
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From:
Marc Chantreux
Date:
June 11, 2019 04:19
Subject:
replace s///r ?
Message ID:
20190608172923.GA3372@prometheus.u-strasbg.fr
hello people,
in perl5, i can
print for
map s/^/4/r,
grep /^\d+$/,
<ARGV>
the perl6 version is a Seq, so much more memory friendly
but i expected (and haven't found in the documentation)
a short equivalent of s///r so the shorter i have is:
$*ARGFILES
. lines
. grep( / ^ <digit>+ $ / )
. map( *.subst(/^/," * ") )
. map(&other-things-to-do)
. map(&say)
when, of course, i want to replace
. map( *.subst(/^/," * ") )
by something like
. map( s:r/^/* / )
any idea?
regards
marc
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replace s///r ?
by Marc Chantreux