What are the reasons for the existence of std::decay
?
In what situations is std::decay
useful?
<joke>It's obviously used to decay radioactive std::atomic
types into non-radioactive ones.</joke>
N2609 is the paper that proposed std::decay
. The paper explains:
Simply put,
decay<T>::type
is the identity type-transformation except if T is an array type or a reference to a function type. In those cases thedecay<T>::type
yields a pointer or a pointer to a function, respectively.
The motivating example is C++03 std::make_pair
:
template <class T1, class T2>
inline pair<T1,T2> make_pair(T1 x, T2 y)
{
return pair<T1,T2>(x, y);
}
which accepted its parameters by value to make string literals work:
std::pair<std::string, int> p = make_pair("foo", 0);
If it accepted its parameters by reference, then T1
will be deduced as an array type, and then constructing a pair<T1, T2>
will be ill-formed.
But obviously this leads to significant inefficiencies. Hence the need for decay
, to apply the set of transformations that occurs when pass-by-value occurs, allowing you to get the efficiency of taking the parameters by reference, but still get the type transformations needed for your code to work with string literals, array types, function types and the like:
template <class T1, class T2>
inline pair< typename decay<T1>::type, typename decay<T2>::type >
make_pair(T1&& x, T2&& y)
{
return pair< typename decay<T1>::type,
typename decay<T2>::type >(std::forward<T1>(x),
std::forward<T2>(y));
}
Note: this is not the actual C++11 make_pair
implementation - the C++11 make_pair
also unwraps std::reference_wrapper
s.
When dealing with template functions that take parameters of a template type, you often have universal parameters. Universal parameters are almost always references of one sort or another. They're also const-volatile qualified. As such, most type traits don't work on them as you'd expect:
template<class T>
void func(T&& param) {
if (std::is_same<T,int>::value)
std::cout << "param is an int\n";
else
std::cout << "param is not an int\n";
}
int main() {
int three = 3;
func(three); //prints "param is not an int"!!!!
}
http://coliru.stacked-crooked.com/a/24476e60bd906bed
The solution here is to use std::decay
:
template<class T>
void func(T&& param) {
if (std::is_same<typename std::decay<T>::type,int>::value)
std::cout << "param is an int\n";
else
std::cout << "param is not an int\n";
}
来源:https://stackoverflow.com/questions/25732386/what-is-stddecay-and-when-it-should-be-used