I've found the following code in Perl 5.10.0's testsuite in t/op/local.t:

@a = ('a', 'b', 'c');
{
    local($a[1]) = 'foo';
    local($a[2]) = $a[2];
    is($a[1], 'foo');
    is($a[2], 'c');
    undef @a;
}

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Now, I daresay (and I hope that my head is clearer than when I wrote my last node) that this only works if "=" evaluates its right-hand operand first.

The important line is the one with the second local. As a run-time operator, local() effectively sets its operand ($a[2] in this case) to an undefined value. But then this new value will be set to the value of $a[2] before local was executed. This would not work the other way round.

Apart from this extract, local.t contains more similar tests.

In a nutshell: Even though I could not find any hints in the documentation, the test suite suggests that kyle's coworker's code is reliable.

In reply to Re: Will "$_[0]=shift" always resolve shift first? by betterworld
in thread Will "$_[0]=shift" always resolve shift first? by kyle

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