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;

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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 ++

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but the buffer is still living as a private member in my object.

In reply to Re: Re: How to swap scalar values without copies by Anonymous Monk
in thread How to swap scalar values without copies by Anonymous Monk

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