c++
Can somebody explain me how has_trivial_default_constructor
works? I tried to find it in a boost implementation but unfortunately there are too many macros and I just got lost...
How somebody can detect the trivial_default_constructor
in C++ using templates?
I need an example in C++ 03 not 11.
#include <boost/type_traits.hpp>
#include <boost/static_assert.hpp>
struct A{
A(){}
int a;
//std::vector< int > b;
};
int main(int argc, char* argv[])
{
struct B{
std::vector< int > b;
};
bool result = boost::has_trivial_default_constructor<A>::value;
//std::forward(&test);
return 0;
}
Actually, it's impossible (in pure C++).
Detecting whether a type has a default constructor is possible with SFINAE because it only involves the interface, but detecting whether it's trivial involves the implementation.
Therefore, a compiler must provide a specific intrinsic for this. You can find a list of intrinsics here, note that some are provided because they require a compiler intervention and others may be provided just to have a uniform set or streamline the Standard Library implementation.
The particular intrinsic you are looking for is either __has_trivial_constructor
which is also supported by gcc and MSVC according to the comments or __is_trivially_constructible
(Clang's specific). I must admit being slightly unsure of the former (what if the type has several constructors ?), the latter can be used as:
template <typename T>
struct has_trivial_default_constructor {
static bool const value = __is_trivially_constructible(T);
};
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