你传递双重 1.1 到非const引用 T& 。这意味着你必须向构造函数传递一个有效的左值，例如：

You're passing the double 1.1 to a non-const reference T&. This means you'd have to pass a valid lvalue to the constructor such as by doing:

double x = 4.3;
test<double> d(x);

使构造函数接受const引用（ const T& ）和它的工作原理，因为你被允许绑定临时（rvalues）到const引用，4.3技术上是一个临时双。

Make the constructor take a const reference (const T&) and it works, because you are allowed to bind temporaries (rvalues) to const references, and 4.3 is technically a temporary double.