C++ Qualification conversions - constness and templates
In the code below, I'm trying to figure out why the compiler (msdev c++ 2010 and Comeau) does not consider the return type of the non specialized template get function to be const. I would expect CASE #2 (see code snipet) to NOT compile, yet it does. Any ideas or links?
Thank you, regu
template < typename T >
struct constness
{
T value;
constness() : value(0) {}
const T &get() { return value; }
};
template < typename T >
struct constness< T * >
{
T * const value;开发者_如何学Go
constness() : value(0) {}
const T * const &get() { return value; }
};
int main( int argc, const char* argv[] )
{
// Uses specialized
constness< double * > wConstness;
const_cast< double * & >(wConstness.value) = new double(1);
*wConstness.get() = 12.0; // CASE #1 doesn't compile
// Uses non specialized
constness< double * const > wConstness2;
const_cast< double * & >(wConstness2.value) = new double(1);
*wConstness2.get() = 12.0; // CASE #2 compiles, allowing modification of
// value pointed by wConstness2.value
return 0;
};
You instantiate constness
with T
= double * const
(a const-qualified pointer to a non-const-qualified double
).
In const T& get()
, the const
applies to the whole T
. If T
was int*
, then const T
would be int* const
, not const int*
. So in your specific case, the const
has no effect because T
is already const-qualified (it is a double* const
).
When you have a typename like T
, adding const
or volatile
to the typename always applies at the top level regardless of what the type is. So, const T
and T const
are always the same, regardless what T
is.
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