Re: What's wrong with this local() ?

in reply to What's wrong with this local() ?

Based on the question, I think you were expecting the localized version to be initialized to the external value. I ran a quick test under Perl v5.8.5, and found that

local $A::variable = $A::variable;
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gives the desired behavior. It's nice, but it doesn't have to be that way, because this behavior implies that the "local" was executed after the RHS of the assignment was evaluated.

Having said that, I don't think that precidence relationship is clearly defined in the language, and you probably shouldn't trust it for production code. Unless someone older and wiser than I knows for sure that it was planned that way?

Comment on Re: What's wrong with this local() ? Download Code
Re^2: What's wrong with this local() ? by Anonymous Monk on Jan 31, 2008 at 17:52 UTC
I think it may have been designed that way. When I look at the operator precedence table, I notice that the '=' is two spots higher in the table than list operators when viewed from the right (local). I also see that '=' evaluates it's right side 1st. So this seems to be a case of behavior matching documentation.	[reply]
Re^3: What's wrong with this local() ? by ikegami (Pope) on Jan 31, 2008 at 18:37 UTC
When I look at the operator precedence table, I notice that the '=' is two spots higher in the table Precedence is irrelevant here. And `local` has higher precedence. Read more... (730 Bytes) It wouldn't work otherwise `Give local higher precedence >perl -le"$x=2; { (local $x)=$x; print $x; $x=3; print $x } print $x" 2 3 2 Give assignment higher precedence >perl -le"$x=2; { local ($x=$x); print $x; $x=3; print $x } print $x" 2 3 3 <-- XXX` [download] I also see that '=' evaluates it's right side 1st. Where did you see this? Operand evaluation order determines which operand is evaluated first. The operand evaluation order for `=` and other Perl binary operators is not documented anywhere and should probably be considered undefined. Perhaps you looked up the associativity of the `=` operator. Associativity doesn't define which operand is evaluated first, it determines which operator is evaluated first when two operators of the same precedence are chained. However, you are right that this feature of `my` and `local` is implemented by choosing a right to left operand evaluation order for assignment operators. `>perl -le"local(${print('a'), 'x'}) = ${print('b'), 'x'}; b a` [download] Updated: Updated code into "precedence is irrelevant" reply.	[reply] [d/l] [select]
Re^2: What's wrong with this local() ? by ikegami (Pope) on Jan 31, 2008 at 18:02 UTC
By design, variables declared with `my` are not introduced (not visible) until after the current statement. For `my`, this is documented in perlsub along with the rest of `my`'s documentation. Similar behaviour exists for `local`, although I can't find it documented.	[reply] [d/l] [select]