C ++模板比int

C ++模板比int-> unsigned转换好吗？(C++ template better than a int->unsigned conversion?)

我有两个功能，如下所示

template<typename T> unsigned int myFunction(T myelement) { myelement->func(); return 2; } void myFunction(unsigned int myelement) { }

我正在使用以下代码

myFunction(2);

visual studio 2012抱怨说“int has not - > func（）”。 为什么不使用unsigned int版本？

I have two functions like the following

template<typename T> unsigned int myFunction(T myelement) { myelement->func(); return 2; } void myFunction(unsigned int myelement) { }

and I'm using the following code

myFunction(2);

visual studio 2012 is complaining that "int hasn't ->func()". Why isn't it using the unsigned int version?

最满意答案

您误读了错误消息。 编译器不使用该函数，它实例化它以确定它是否是候选者。 您需要为不合适的类型禁用实例化：

template<typename T> typename std::enable_if<!std::is_fundamental<T>::value, unsigned int >::type myFunction( T myelement ) { // ... }

实例

You are misreading the error message. The compiler doesn't use the function, it is instantiating it to figure out if it is a candidate. You need to disable the instantiation for non-suitable types:

template<typename T> typename std::enable_if<!std::is_fundamental<T>::value, unsigned int >::type myFunction( T myelement ) { // ... }

Live example