I looked into the GCC STL (4.6.1) and saw that std::copy()
uses an optimized version in case the builtin __is_trivial()
evaluates to true
.
Since the std::copy()
and std::reverse_copy()
templates are very useful for copying elements in arrays, I would like to use them. However, I have some types (which are results of template instantiations) that are structs that contain some trivial values, no pointers and have no copy constructor or assignment operator.
Is G++ smart enough to figure out that my type in fact is trivial? Is there any way in C++98 to make sure an STL implementation knows that my type is trivial?
I guess that in C++11, things will become convenient using the is_trivial<>
type trait. Is this right?
Thanks!
EDIT: Sorry for being so late with this, but here is an example of a pretty simple Type
class that is not trivial to GCC and llvm. Any ideas?
#include <iostream>
struct Spec;
template <typename TValue, typename TSpec>
class Type
{
public:
TValue value;
Type() : value(0) {}
};
int main()
{
std::cerr << "__is_trivial(...) == "
<< __is_trivial(Type<char, Spec>) << '\n';
return 0;
}
__is_trivial
gives the right value for all types.You can't rely on any paticular STL implementation to use it, although if the compiler vendor supplied it then it probably does contain various compiler-specific improvements behinds the scenes.
C++11's
std::is_trivial
is merely standardising this feature, there's no reason for an implementation not to use it.There has been some debate over what
trivial
meant.For example, your example is not trivially constructible as far as I can tell (
std::is_trivially_default_constructible
would return false I think).In your case, I think you would need the new trait
std::is_trivially_copyable
, which is more fine grained. So... upgrade your compiler ?