Re: Re: Short Circuit Operator and Hash Assignment

The 1 && part is being optimized out of the statement.

this seems a rather dangerous and naive optimization to me, given that also this:

$x = 1 && $y = 2;
$x = 0 || $y = 2;
[download]

gets "optimized" as:

$x = $y = 2;
$x = $y = 2;
[download]

perhaps the p5p people should do something about it.

cheers,
Aldo

King of Laziness, Wizard of Impatience, Lord of Hubris

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Re: Short Circuit Operator and Hash Assignment by jonadab (Parson) on Mar 17, 2004 at 18:24 UTC
this seems a rather dangerous and naive optimization to me The optimization is a red herring; it's not changing the result in any way. If the left operand of the `&&` operator is true in boolean context, then it returns the value of its right operand, whatever that may be. Expecting it to return the left operand is wrong. (What you really were expecting is a different parse, based on different operator precedence. (See my other post downthread.)) `;$;=sub{$/};@;=map{my($a,$b)=($_,$;);$;=sub{$a.$b->()}}` `split//,".rekcah lreP rehtona tsuJ";$\=$;[-1]->();print`	[reply] [d/l]
Re: Re: Re: Short Circuit Operator and Hash Assignment by hardburn (Abbot) on Mar 17, 2004 at 17:33 UTC
You're hitting an edge case. The optimization had a different situation in mind: `if( 1 && foo() ) { . . . } elsif( 0 \|\| bar() ) { . . . }` [download] This optimization is hardly unique to Perl. I believe both C and Java compilers will optimize away constants in exactly the same way. ---- `: () { :\|:& };:` Note: All code is untested, unless otherwise stated	[reply] [d/l] [select]