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Re^4: Boolean counter?

by ikegami (Pope)
on Dec 07, 2009 at 16:09 UTC ( #811530=note: print w/ replies, xml ) Need Help??


in reply to Re^3: Boolean counter?
in thread Boolean counter?

What about it is not defined?


Comment on Re^4: Boolean counter?
Re^5: Boolean counter?
by ikegami (Pope) on Dec 07, 2009 at 16:57 UTC

    duh, you're referring to the fact that he's assigning to the variable he just modified. I was still asleep.

    While such constructs should be avoided over clarity concerns, it's wrong to say it's undefined. The precedence and evaluation order for that statement is actually clearly defined in perlop.

      Quoting perlop:
      Note that just as in C, Perl doesn’t define when the variable is incremented or decremented. You just know it will be done sometime before or after the value is returned. This also means that modifying +a variable twice in the same statement will lead to undefined behaviour. Avoid statements like: $i = $i ++; print ++ $i + $i ++; Perl will not guarantee what the result of the above statements is.
      To me, that seems that perlop quite explicitly says the behaviour is undefined.

      Perhaps you have an other reading of perlop which defines the order?

        You're countering a statement about a specific Perl statement with statement about the general case.

        Yes, operand evaluation order is officially not defined. Assignments are a special case. Assignments are guaranteed to evaluate their RHS argument first to allow the following to work:

        my $i = $i; local $i = $i;

        It does seem to be somewhere other than perlop.

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