I am saying that, with defined EO, that func1() and func2() can be parallelised. 

my $var = func1($a) + func2($a + 1);

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You say "compare this with...The compiler is free to do this in any order it pleases" and then build the rest of you arguement upon that basis.

But with defined execution order, the compiler is not free to "do this is any order".

From that point on, you build sandcastles in the air based on that false premise.

Update: Think about what makes a routine parallelisable? Not just what could be parallelised, buit what would make parallelisation worth while. ANd how the compiler would know that.

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?

In reply to Re^32: Why is the execution order of subexpressions undefined? (magic ruts) by BrowserUk
in thread Why is the execution order of subexpressions undefined? by BrowserUk

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