在 2019-03-25一的 09:32 +0800，叶佑群写道：
> Hi,all, I have code as below:
>
> void check(int & val)
> {
> std::cout << "int &\n";
> }
>
> void check(int && val)
> {
> std::cout << "int &&\n";
> }
>
> template <typename T>
> void print (T && val)
> {
> check (std::forward<T>(val));
> }
>
> int main()
> {
> int x = 0;
> print (x);
> print (std::move (x));
> }
>
> I have some problems:
>
> A、when call print (std::move (x)), the compiler will deduction print to
> Void print (int && val)
>
> So val is a rvalue reference, so check (int && val) should be
> invoked, but actually not, why? Function forward just do a static_cast here,
> its result also make val a rvalue reference?
I can not understand your problem so I'll just explain what is happening.
When you call "print(std::move(x))", "T" is deducted to be "int". When you call
"print(x)", "T" is deducted to be "int &". So "T &&" is "int &" (reference
collapsing). Let's instantiate them:
void print<int>(int &&val)
{
check(std::forward<int>(val));
}
void print<int &>(int &val)
{
check(std::forward<int &>(val));
}
std::forward is defined as-is:
template <class T> constexpr T&& forward(remove_reference_t<T>& t) noexcept
{
return static_cast<T&&>(t);
}
template <class T> constexpr T&& forward(remove_reference_t<T>&& t) noexcept
{
return static_cast<T&&>(t);
}
In print<int> it will be instantiated:
constexpr int&& forward<int>(remove_reference_t<int>&& t) noexcept
{
return static_cast<int&&>(t);
}
In print<int&>, reference collapsing happens again since "T" is "int &", so
"T&&" is "int &":
constexpr int& forward<int&>(remove_reference_t<int&>& t) noexcept
{
return static_cast<int&>(t);
}
--
Xi Ruoyao <xry111@xxxxxxxxxxxxxxxx>
School of Aerospace Science and Technology, Xidian University
