[英]How reference deduce works?
May be duplicated to this . 可以复制到此 。
I read Effective Modern C++ . 我读了《 有效的现代C ++》 。 Under Item 1
, I found a case for universal reference: 在Item 1
下,我找到了一个可供普遍参考的案例:
For the last example f(27);
对于最后一个例子, f(27);
I did a test under VS2013. 我在VS2013下进行了测试。
void process(int& x)
{
std::cout << "int&" << std::endl;
}
void process(int&& x)
{
std::cout << "int&&" << std::endl;
}
template<typename T>
void f(T&& param)
{
std::cout << "------------------------------------------------" << std::endl;
if (std::is_lvalue_reference<T>::value)
{
std::cout << "T is lvalue reference" << std::endl;
}
else if (std::is_rvalue_reference<T>::value)
{
std::cout << "T is rvalue reference" << std::endl;
}
else
{
std::cout << "T is NOT lvalue reference" << std::endl;
}
std::cout << "param is: " << typeid(param).name() << std::endl;
process(std::forward<T>(param));
process(param);
}
int getINT()
{
int x = 10;
return x;
}
int _tmain(int argc, _TCHAR* argv[])
{
f(10);
f(getINT());
return 0;
}
Here is the output: 这是输出:
------------------------------------------------
T is NOT lvalue reference
param is: int
int&&
int&
------------------------------------------------
T is NOT lvalue reference
param is: int
int&&
int&
I found that within the template function, without std::forward<T>(param)
, process(int& x)
will be called but according to the book, the type for param
should be rvalue reference, so process(int&& x)
should be called. 我发现在没有std::forward<T>(param)
的模板函数中,将调用process(int& x)
但根据本书, param
的类型应为右值引用,因此process(int&& x)
应为叫做。 But this is not the case. 但这种情况并非如此。 Is it I misunderstand something? 我误会了吗?
Here is the forwarding reference I found from other thread : 这是我从其他线程找到的转发参考:
You're confusing types with value categories . 您正在将类型与值类别混淆。 As a named parameter, param
is an lvalue, then for process(param);
作为命名参数, param
是一个左值,然后对于process(param);
process(int& x)
will be called. process(int& x)
将被调用。
That's why we should use std::forward
with forwarding reference ; 这就是为什么我们应该在转发参考中使用std::forward
; std::forward<T>(param)
converts param
to rvalue for this case, then process(int&& x)
will be called (as expected). 在这种情况下, std::forward<T>(param)
将param
转换为rvalue,然后将调用process(int&& x)
(如预期的那样)。
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