Re: A non-const reference may only be bound to an lvalue?

From:
David Wilkinson <no-reply@effisols.com>
Newsgroups:
microsoft.public.vc.language
Date:
Mon, 17 Dec 2007 09:01:56 -0500
Message-ID:
<Ori$gVLQIHA.5360@TK2MSFTNGP03.phx.gbl>
George wrote:

Hi Abhishek,

I have some difficulties to understand below code about how it is executed,

return const_cast<T>( static_cast<const std::vector<T>> (vec)[i]);


1. It first converts vec to vector<T>?

static_cast<const std::vector<T>> (vec)

2. then gets its ith element?

[i]

3. finally remove const qualification on T itself?? I am confused. T is a
type, not a variable?


George:

Actually, I think the example is not quite presented correctly. I
believe the idea behind it is to eliminate duplication (which might be
more important in a more complex example).

How about this:

#include<vector>
#include <assert.h>

template<typename T>
class A
{
    std::vector<T> vec;
public:
    explicit A(size_t n = 0):
    vec(n)
    {
    }

    const T& operator[](size_t i) const
    {
      return vec[i];
    }

    T& operator[](size_t i)
    {
      return const_cast<T&>(operator[](i));
    }
};

The non-const version is defined in terms of the const one, eliminating
duplication.

int main()
{
   A<double> a(1);
   double& x = a[0]; // uses non-const version
   assert(x == 0.0);
   return 0;
}

--
David Wilkinson
Visual C++ MVP

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