Re: const correctness - should C++ prefer const member over non-const?

From:
blargg.h4g@gishpuppy.com (blargg)
Newsgroups:
comp.lang.c++
Date:
Thu, 30 Oct 2008 05:41:31 -0500
Message-ID:
<blargg.h4g-3010080541310001@192.168.1.4>
In article
<8141ce63-f45b-4692-a05b-e6c44b367c7c@e17g2000hsg.googlegroups.com>,
fungus <openglMYSOCKS@artlum.com> wrote:

On Oct 30, 10:21 am, fungus <openglMYSO...@artlum.com> wrote:

Ok, maybe I oversimplified it. Supposed operator[]
returns a reference to the int:


...and just before the pedants arrive, suppose it's
a struct not an int, and I want to access a member
of the stuct.

my_struct& operator[](int n) { return data[n]; }
const my_struct& operator[](int n) const { return data[n]; }

Why does the compiler choose the non-const version
for the RHS of an expression...?


That's a better question. It chooses the non-const version because C++
doesn't overload based on return type or how the caller uses the return
value, only arguments to the function (including the implicit "this"
argument to member functions). I could have sworn "The Design and
Evolution of C++" covered the reason behind this, but I couldn't find a
reference. I'm assuming it would complicate overloading and often not be
desired.

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