模板函數自動返回類型與std :: conditional

Question

我想編寫一個可以采用的通用函數

（1）A的右值引用並返回移動構造的類型B：

A a;
B b = f(std::move(A));
// use b normally and a is now empty.

或（2）A的左值引用並返回一個包裝自動類型扣除A的包裝器對象：

A a;
auto b = f(A);
// a could be used normally. The type of b is of a very complex form deduced based on the implementation of f()
B bb = b.eval(); // use eval() method from the auto type to make an evaluation of the wrapper object b to effectively copy-construct bb from a. 

我可以通過執行以下操作來完成此操作：

template <typename T>
auto f(T&& t)
 -> typename std::conditional
          <!std::is_lvalue_reference<T>::value,
           T,
           decltype (t.sqrt() + (t * 2).sin())  // ideally this could be decltype(auto) to not repeat the code but this is not allowed by C++
          >::type
{
    T _t(std::forward<T>(t));
    return _t.sqrt() + (_t * 2).sin()  // T is a data type of some template expression library with lazy evaluation capability, e.g., Eigen::Matrix, hence this line will return an expression type that is composed from the types involved in the expression, i.e. sqrt, +, *, sin. 
 }

我的問題是，正如上面代碼的注釋中所指出的，如何在不使用decltype(auto)作為auto關鍵字的情況下在decltype()調用中刪除重復計算，這在std::conditional模板參數中是禁止的？

先感謝您！

Answer 1

沒有重復計算，只是代碼重復。 使用函數可以避免代碼重復。

template <typename T>
auto f(T&& t)
 -> typename std::conditional
          <!std::is_lvalue_reference<T>::value,
           T,
           decltype (function(t))
          >::type
{
    T _t(std::forward<T>(t));
    return function(_t);
}