Re: How to swap scalar values without copies

My curiosity has finally got the best of me, and I can no longer resist asking (though I really tried to let it go) why references are out of the question.

As an OO programmer you're certanly comfortable with references already, so there must be a good reason why they can't be used in this case.

I'm unable to restrain myself from posing the question. I hope I'm about to learn something new. ;)

Dave

Comment on Re: How to swap scalar values without copies

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Re: Re: How to swap scalar values without copies by Anonymous Monk on Feb 20, 2004 at 18:25 UTC
Well, I'm currently confortable with C++ references. Although the name is the same, the syntax is different. In C++ I can write: `int &pluto = pippo;` [download] and, from now on (up to the end of its scope), `pluto` is just a synonim for `pippo`. I'am not asserting I can do this for an object member. In C++, if I have a class member named `data`, in a method of that class I can simply refer to it as `data`, which is the same as `this->data`. In Perl, the same member must be referred to as `$this->{data}`, in addition to explicitely setting the object pointer every time: `my $this = shift;`, for instance. But, if I store a reference, I am obliged to refer to the corresponding scalar as `${$this->{data}}`, and this clutters the code way too much for me (`$$data` would be acceptable) . Please, tell me that I am wrong and I can save all this typing. Moreover, just saving a reference to an argument as a class member is not completely satisfactory. The top of my dreams would be that the passed buffer becomes a private member, i.e. it is not possible to modify it from outside the constructor by referring to the actual argument: `read(IN, $buffer, 1000000); push @mysegments, new MyPackage::MySegment($buffer); print "+$buffer+"; # gives ++` [download] but the buffer is still living as a private member in my object.	[reply] [d/l] [select]

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Re: Re: How to swap scalar values without copies
by Anonymous Monk on Feb 20, 2004 at 18:25 UTC

int &pluto = pippo;
[download]

pluto

pippo

In C++, if I have a class member named data, in a method of that class I can simply refer to it as data, which is the same as this->data. In Perl, the same member must be referred to as $this->{data}, in addition to explicitely setting the object pointer every time: my $this = shift;, for instance.

But, if I store a reference, I am obliged to refer to the corresponding scalar as ${$this->{data}}, and this clutters the code way too much for me ($$data would be acceptable) . Please, tell me that I am wrong and I can save all this typing.

Moreover, just saving a reference to an argument as a class member is not completely satisfactory. The top of my dreams would be that the passed buffer becomes a private member, i.e. it is not possible to modify it from outside the constructor by referring to the actual argument:

read(IN, $buffer, 1000000);
push @mysegments, new MyPackage::MySegment($buffer);
print "+$buffer+"; # gives ++
[download]

[reply]
[d/l]
[select]