Re^3: Without trying; what do you think this would print? (/left/ side)

You're right tye, it's a bug. The following should be equivalent:

$ perl -e '$a{0} = $a{0} || scalar keys %a; print $a{0}'
0
$ perl -e '$a{0} ||= scalar keys %a; print $a{0}'
1
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Tested on 5.8, 5.10, and 5.12

Comment on Re^3: Without trying; what do you think this would print? (/left/ side) Download Code

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Re^4: Without trying; what do you think this would print? (notabug!) by tye (Sage) on Apr 30, 2011 at 15:49 UTC
No, it isn't. No, they need not be. - tye	[reply]
Re^5: Without trying; what do you think this would print? (notabug!) by wind (Priest) on Apr 30, 2011 at 18:06 UTC
Well, from my basic understanding of perlop - Assignment Operators those two should be equivalent. Maybe I'm wrong though. Fortunately, it's a contrived example anyway.	[reply]
Re^6: Without trying; what do you think this would print? (notabug!) by tye (Sage) on Apr 30, 2011 at 19:30 UTC
Yes. They are both even equivalent to the point of either producing 0 or 1 depending on implementation details including optimizations. - tye	[reply]
Re^7: Without trying; what do you think this would print? (notabug!) by wind (Priest) on Apr 30, 2011 at 20:39 UTC
Re^8: Without trying; what do you think this would print? (notabug!) by tye (Sage) on Apr 30, 2011 at 22:18 UTC
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