Re^6: Why is the execution order of subexpressions undefined? (sane?)

So does that mean $z is calculated before $y+1 or not? In what order does $w get calculated?

$w= $z;
$x= ( $a + $w ) * ( $y + 1 ) + ( $a + $z ) / ( $y + 1 );
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The optimizer notices that $w and $z must be equal and so we need only add $a to their shared value once. In what order does this single operation happen relative to the rest of the calculations? If I change the first assignment, surely the order of operations for the second assignment doesn't change. What is the obvious answer to this?

- tye

Comment on Re^6: Why is the execution order of subexpressions undefined? (sane?) Download Code

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Re^7: Why is the execution order of subexpressions undefined? (sane?) by BrowserUk (Patriarch) on Apr 12, 2005 at 06:44 UTC
The optimizer notices that $w and $z must be equal... A C or C++ compiler might be able to make that optimisation, but Perl--No way! What if $w is tied? What if `my Bit2 $w;` and `my Bigint $z;`? Bogus question. Next? Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. Lingua non convalesco, consenesco et abolesco. Rule 1 has a caveat! -- Who broke the cabal?	[reply] [d/l] [select]
Re^8: Why is the execution order of subexpressions undefined? (sane?) by theorbtwo (Prior) on Apr 18, 2005 at 12:33 UTC
Note that tye mentioned perl6, which will allow you to mention the types of your variables, so that the compiler /can/ tell. Warning: Unless otherwise stated, code is untested. Do not use without understanding. Code is posted in the hopes it is useful, but without warranty. All copyrights are relinquished into the public domain unless otherwise stated. I am not an angel. I am capable of error, and err on a fairly regular basis. If I made a mistake, please let me know (such as by replying to this node).	[reply]