Today in "use overload", the so-called assignment operator can be defined. If an overloaded operator is called on an object that is not unique but shares a reference, this function is called to duplicate it. That is, it's uses as a lazy copy-on-write mechanism.

This means that Perl must know that the reference is shared, and which functions are indeed "mutators". I can have a complex number class appear to have value semantics when an assignment is followed by an overloaded operator such as "+", but not for a named method call.

So, something like 

$a= $b;
++ $a;

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can indeed be made to work like built-in numbers, calling 
$a= $b;
$a->set_imaginary (5);

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cannot, and would set the imaginary part of $b as well.

Perhaps Perl6 simply needs to have the lazy copy function applied uniformly whenever a mutator is called, whether you must identify mutators or it figures it out by itself.

Meanwhile, why write the copy code every time? Simply saying that a class is "by value" should be enough to automatically generate the copy function, which copies all the data members of the object.

—John

In reply to Re: Re: Perl6 - value vs. reference issues by John M. Dlugosz
in thread Perl6 - value vs. reference issues by John M. Dlugosz

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