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

From:

Abhishek Padmanabh <abhishek.padmanabh@gmail.com>

Newsgroups:

microsoft.public.vc.language

Date:

Mon, 17 Dec 2007 06:07:02 -0800 (PST)

Message-ID:

<aadc9aac-4cac-4bcb-ab23-85c071fc33b0@b40g2000prf.googlegroups.com>

On Dec 17, 7:01 pm, David Wilkinson <no-re...@effisols.com> wrote:

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));
    }

This, I believe is incorrect. It will cause infinite recursion, won't
it? You need to const-ify the this pointer to invoke the const member
functions.