[erlang-questions] Reading, Learning, Confused

Edwin Fine erlang-questions_efine@REDACTED
Sat Jul 19 19:40:40 CEST 2008


I am not sure I understand what is "not quite right". All I can see is that
I didn't mention that a failed (as in, there was an exception or error)
guard condition is treated as false.

I think your code works exactly as I would have expected. I see the sequence

First Clause:
f(0) when (0 == 0) [true] or ((1/X) > 2) [error = false] => entire guard
fails, so try next clause.
Next Clause:
f(0) when (0 == 0) true => short-circuit all subsequent guard conditions, so
guard is true and clause is executed.

I was saying that the absence of side-effects in guard conditions allows
Erlang to short-circuit  guard expressions of the format

   when cond1; cond2; cond3


   when cond1, cond2, cond3

This was not referring to use of or/orelse/and/andalso, which I tend to
avoid in guards.

Please help me understand what you meant in your post.

On Sat, Jul 19, 2008 at 12:59 PM, Darren New <dnew@REDACTED> wrote:

> Edwin Fine wrote:
> > of guards, it is guaranteed that leaving out the evaluation of one or
> > more guards will not change the state of the program.
> I don't think that's quite right.
> f(X) when (X == 0) or ((1 / X) > 2) ->
>    "does not return for zero X";
> f(X) when (X == 0) orelse ((1 / X) > 2) ->
>   "does return for zero X".
> I might be misunderstanding here, but I understand that an "abrupt
> return" from a calculation in a guard is treated the same as "false".
> So in the first case, when X is 0, the guard evaluates to false, because
> 1/X errors out, making the entire expression false.
> In the second case, 1/X isn't evaluated when X==0, making the entire
> expression true.
> It does seem like the documentation is wrong. I just tried the above
> fragment and it printed "does return" as its answer.
> --
> Darren New / San Diego, CA, USA (PST)
>  Helpful housekeeping hints:
>   Check your feather pillows for holes
>    before putting them in the washing machine.
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