Re: A non-const reference may only be bound to an lvalue?
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:
explicit A(size_t n = 0):
const T& operator(size_t i) const
T& operator(size_t i)
The non-const version is defined in terms of the const one, eliminating
double& x = a; // uses non-const version
assert(x == 0.0);
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