Re^2: Strange aliasing(?) when modifying variable in range operator

Thank you for answer, Lanx, ~~but range operator is not a "call" (to function), is it? And operators aren't (shouldn't be(?)) executed before arguments are evaluated?~~

Hm-m, I'm wrong. Back to school...

>perl -wE "$i=1; say $i + ($i+=1)"
4
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Comment on Re^2: Strange aliasing(?) when modifying variable in range operator Download Code

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Re^3: Strange aliasing(?) when modifying variable in range operator by LanX (Saint) on Jun 09, 2018 at 18:19 UTC
You're welcome! :) NB: You have no guaranty at all when INSIDE the statement the increment happens, it's only guarantied AFTER the statement. Most probably is Perl doing internal optimizations here. Your do-block examples puzzle me, please note the difference between debugger and one-liner `DB<1> $i=1; say for do{$i?$i:($i+1)}..$i++ #6 1 DB<2> q D:\Users\lanx>perl -wE "$i=1; say for do{$i?$i:($i+1)}..$i++ #6" D:\Users\lanx>` [download] again, unpredictable internal optimization. If interested, you could use B::Concise to track what is happening here... So Rule Of Thumb: NEVER mix $i and $i++ (or --$i) inside the same statement. Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l]

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Re^3: Strange aliasing(?) when modifying variable in range operator
by LanX (Saint) on Jun 09, 2018 at 18:19 UTC

NB: You have no guaranty at all when INSIDE the statement the increment happens, it's only guarantied AFTER the statement.

Most probably is Perl doing internal optimizations here.

Your do-block examples puzzle me, please note the difference between debugger and one-liner

  DB<1> $i=1; say for do{$i?$i:($i+1)}..$i++   #6
1


  DB<2> q

D:\Users\lanx>perl -wE "$i=1; say for do{$i?$i:($i+1)}..$i++   #6"

D:\Users\lanx>
[download]

again, unpredictable internal optimization.

If interested, you could use B::Concise to track what is happening here...

So Rule Of Thumb:

NEVER mix $i and $i++ (or --$i) inside the same statement.

Cheers Rolf
_{(addicted to the Perl Programming Language :)

Wikisyntax for the Monastery}

[reply]
[d/l]