Re: Order of evaluation/interpolation of references (alias vs copy)

Perhaps because "${X()}${X()}" does ${X()} . ${X()} before any stringification makes a copy, so both aliases to the hidden $x are incremented again before their values are copied to the new string. While the other case turns into ( ${X()} . ' ' ) . ${X()} so the first alias to the hidden $x gets copied into a string with a trailing space before $x gets incremented again.

Lots of surprises to discover when you do such things.

- tye

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Re^2: Order of evaluation/interpolation of references (alias vs copy) by LanX (Saint) on Mar 07, 2012 at 21:49 UTC
I think you're right! `lanx:/tmp$ perl -MO=Deparse,-p,-q -e 'print "${X()} ${X()}\n"' print((((${X();} . ' ') . ${X();}) . "\n")); lanx:/tmp$ perl -MO=Deparse,-p,-q -e 'print "${X()}${X()}\n"' print(((${X();} . ${X();}) . "\n"));` [download] Cheers Rolf	[reply] [d/l]

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Re^2: Order of evaluation/interpolation of references (alias vs copy)
by LanX (Saint) on Mar 07, 2012 at 21:49 UTC

lanx:/tmp$ perl -MO=Deparse,-p,-q -e 'print "${X()} ${X()}\n"'
print((((${X();} . ' ') . ${X();}) . "\n"));
    
lanx:/tmp$ perl -MO=Deparse,-p,-q -e 'print "${X()}${X()}\n"'
print(((${X();} . ${X();}) . "\n"));
[download]

Cheers Rolf

[reply]
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