为什么使用std :: forward禁用模板参数扣除? [英] Why is template argument deduction disabled with std::forward?
问题描述
在VS2010中,std :: forward定义如下:
In VS2010 std::forward is defined as such:
template<class _Ty> inline
_Ty&& forward(typename identity<_Ty>::type& _Arg)
{ // forward _Arg, given explicitly specified type parameter
return ((_Ty&&)_Arg);
}
出现在
仅用于禁用模板参数扣除。
identity
appears to be used solely to disable template argument deduction. What's the point of purposefully disabling it in this case?
推荐答案
如果您将一个右值引用传递给一个类型<$ c的对象$ c> X 到模板函数,该函数以类型 T&&
作为其参数,模板参数扣除推导 T
为 X
。因此,参数具有类型 X&&&
。如果函数参数是一个左值或常量值,则编译器将其类型推断为该类型的左值引用或常量左值引用。
If you pass an rvalue reference to an object of type X
to a template function that takes type T&&
as its parameter, template argument deduction deduces T
to be X
. Therefore, the parameter has type X&&
. If the function argument is an lvalue or const lvalue, the compiler deduces its type to be an lvalue reference or const lvalue reference of that type.
如果 std :: forward
使用模板参数扣除:
If std::forward
used template argument deduction:
由于对象的名称为lvalue
,只有 std :: forward
正确地转换为 T&&&
将是输入参数是未命名的右值(如 7
或 func()
)。在完美转发的情况下,传递到 std :: forward
的 arg
是一个左值,因为它有一个名称。 std :: forward
的类型将被推导为一个左值引用或常量值引用。参考折叠规则会导致std中的 T&&< / code> in
static_cast< T&&>(arg)
:: forward to always resolve as a lvalue reference or const lvalue reference。
Since objects with names are lvalues
the only time std::forward
would correctly cast to T&&
would be when the input argument was an unnamed rvalue (like 7
or func()
). In the case of perfect forwarding the arg
you pass to std::forward
is an lvalue because it has a name. std::forward
's type would be deduced as an lvalue reference or const lvalue reference. Reference collapsing rules would cause the T&&
in static_cast<T&&>(arg)
in std::forward to always resolve as an lvalue reference or const lvalue reference.
示例:
template<typename T>
T&& forward_with_deduction(T&& obj)
{
return static_cast<T&&>(obj);
}
void test(int&){}
void test(const int&){}
void test(int&&){}
template<typename T>
void perfect_forwarder(T&& obj)
{
test(forward_with_deduction(obj));
}
int main()
{
int x;
const int& y(x);
int&& z = std::move(x);
test(forward_with_deduction(7)); // 7 is an int&&, correctly calls test(int&&)
test(forward_with_deduction(z)); // z is treated as an int&, calls test(int&)
// All the below call test(int&) or test(const int&) because in perfect_forwarder 'obj' is treated as
// an int& or const int& (because it is named) so T in forward_with_deduction is deduced as int&
// or const int&. The T&& in static_cast<T&&>(obj) then collapses to int& or const int& - which is not what
// we want in the bottom two cases.
perfect_forwarder(x);
perfect_forwarder(y);
perfect_forwarder(std::move(x));
perfect_forwarder(std::move(y));
}
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