Sub expressions are evaluated before the expressions they are a part of. Order of evaluation within a subexpression is in precedence order with ties broken left to right.

In the expression you gave, both ( $y + 1 ) are subexpressions and would be evaluated before the overall expression, left to right. The fact that they are the same subexpression gives rise to the possible optimisation.

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        

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?

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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?