Re: Re: Incrementing a Hash Value

[marvell]

Is it undefined, or does do this:

# basic form
$i = $i++;

# long hand
$k = $i;
$i++;
$i = $k;

--
¤ Steve Marvell

[Abigail-II]

It's undefined. See also http://bit.csc.lsu.edu/tutorial/ten-commandments/c-faq/c-4.html (Perl defines the behaviour of ++ as to "work as in C" - as the perlop manual page will tell you).

Abigail

[impossiblerobot]

I'm definitely not qualified to argue whether this behavior should be considered 'undefined', but I haven't seen any pointers to documentation that says that it is.

perlop is written in a language which is a little less than explicit than Perl, but it doesn't say that ++ and -- work exactly as in C.

In fact, perlop says:

"++" and "--" work as in C. That is, if placed before a variable, they increment or decrement the variable before returning the value, and if placed after, increment or decrement the variable after returning the value.

In other words, the auto-increment operator works as in C in the specific way mentioned. (perlop then goes on, in the next paragraph, to explain one way in which the perl implementation differs from C.)


Impossible Robot

[Juerd]

Is it undefined, or does do this: $k = $i; $i++; $i = $k;

Both.

$i++ stores the old value, increases the value, and then sets the return value to the stored old value. Although that is true for all versions of Perl, its behaviour is not documented. Actually, it is documented, but wrong:

[...] increment or decrement the variable after returning the value. [...] -- perlop

It doesn't increment after returning. It increments first, and then returns the old value.

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[marvell]

That sounds well defined to me. If it has a predicible behaviour, even if it's not useful, it's defined.

I must be missing something.

--
Steve Marvell

[Abigail-II]

But it isn't well defined. In

    $i = $i ++;

you do not know whether the assignment will happen before, or after the increment.

Abigail

[Abigail-II]

It doesn't increment after returning. It increments first, and then returns the old value.

No, and that's the entire problem. You do not know when the increment will happen, that is specifically left undefined. (It was left undefined in C to give implementers of compilers all sorts of freedom to optimize it - idem for Perl). What a specific version of Perl does not gives you absolutely no garantee what it will do in a next version.

Abigail

[Juerd]

No, and that's the entire problem. You do not know when the increment will happen, that is specifically left undefined.

It is documented to be undefined (as you can't know when "after" will be), but the documentation is wrong ("before" is reality: the old value is stored in a temporary SV).

(It was left undefined in C to give implementers of compilers all sorts of freedom to optimize it - idem for Perl)

Perl will never be able to optimize $foo++ for real scalars, because they can be tied or overloaded at runtime. With Perl 6, when there are variables that are numbers but can never become a real scalar, maybe it can be optimized.

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