in reply to Re: Short Circuit Operator and Hash Assignment
in thread Short Circuit Operator and Hash Assignment

Note, too, that the exact operator precedance in Perl5 is poorly defined, and there are many edge cases that don't work out quite as you'd expect.
Gah, you make it sound as though Perl rolls dice to determine operator precedence. The exact operator precedence in Perl is quite well-defined, and there are no random number generators involved. But people fail to understand it in two different ways. First, they simply don't bother to learn the precedence table, which was the problem in this case (and is the usual problem). I think people get lazy because operator precedence usually DWIMs, so they figure it will always DWIM, and then are surprised when it doesn't.

The other way that people fail to understand operator precedence is that they don't know when it doesn't apply. It's important to understand that a yacc grammar uses operator precedence only to break ties between ambiguous rules in the grammar. If the grammar specifies that something can only be parsed in a particular way, operator precedence never comes into it. You may think that the $ on the front of a variable is a unary operator, but it's not, because what comes after it is not allowed to be a general expression. The grammar restricts the following item to be an identifier, a simple scalar variable, or a block, and these are all recognized by the grammar without recourse to operator precedence. If you try to treat $ as a unary operator, you will certainly be surprised when it doesn't work that way. But it's not an "edge case" of operator precedence. It's simply getting nowhere near the operator precedence table in the first place.

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Re: Re: Re: Short Circuit Operator and Hash Assignment
by hardburn (Abbot) on Mar 17, 2004 at 19:10 UTC

    I was thinking of "edge cases" where there is a tie in the precedence table, such as in Auto-increment frenzy. Which seems to be a recurring theme on the Fun With Perl mailing list.

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      Then it would seem that you either didn't read or didn't understand the thread in question, since in Re: Auto-increment frenzy Abigail-II correctly points out that the order of evaluation in that case has nothing whatsoever to do with operator precedence. Operator precedence has absolutely nothing to say about the order in which the two terms to a binary operator are evaluated. Terms ain't operators. But even if you happen to know or guess that the left term is usually evaluated before the right one, you still can't assume that the post-increment happens right when the old value is returned. Legally it can happen any time up until the next statement. But again, this is a run-time issue, not a parsing issue. Just because operator precedence places some constraints on the order of evaluation doesn't mean that we should use the former term to mean the latter. Okay, maybe you did understand the thread--I can see that maybe you're just using the term differently, but if so, what you have to understand is that "operator precedence" is a technical term in computer science, and to use it generically like that will make computer scientists think you're spouting nonsense.

      So yes, there are certainly "edge cases" in the order of evaluation when it comes to side effects. Every imperative computer language has them. (Which is one reason people fall in love with functional languages.) So in the documentation for an imperative language, you'll almost always find something that says something like: "Relying on the order of evaluation for side effects is erroneous. Just don't do that. Use a separate statement if you want to guarantee order of evaluation."

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        Yes, I re-read that thread about five minutes after the post above. I see what you mean in regards to operator vs. evaluation precedence.

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