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

David Wilkinson <>
Mon, 17 Dec 2007 09:01:56 -0500
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?


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


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 <assert.h>

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

    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

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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