As others have explained, this is evaluated left to right, but <coed>$i</code> is passed by reference. You can try something like 

sub copy { my @t = @_ };
my $rv = func( copy($i), copy(++i), copy($i+2) );

[download]
or just 
my $rv = func( 0 + $i, 0 + ++i, $i + 2);

[download]

(This one's tricked me too once, until the monks was kind enough to explain it in the cb.)

In fact, perl evaluates most operators from left to right, and surely always evaluates comma from left to right (whether in scalar or list context), even if this is undocumented. There can still be exceptions, for example an assigment is evaluated rhs first.

While execution order is undocumented in perl, you can count much more on it then in languages where it is actually undefined (liek scheme or C or C++), because there's only one perl interpreter and it's not likely to change much.

There is a certain amount of relegious war between fixed and undefined evaluation order. The arguments are like this. Some say that undefined evaluation order is better because it can be optimized better. Others say that this is non-sense, since the compiler can often determine that simple sub-expressions don't have side effects, and thus evaluate them in a different order anyway; and when the subexpression do have side effects, undefined evaluation order can cause confusion and misterious bugs. There's a point in both arguments, and I'm not sure which one I support.

In reply to Re: Why is the execution order of subexpressions undefined? by ambrus
in thread Why is the execution order of subexpressions undefined? by BrowserUk

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