为什么编译器选择此模板函数而不是重载的非模板函数? [英] Why is the compiler choosing this template function over an overloaded non-template function?

问题描述

使用VC ++ 2010，具体如下:

Using VC++ 2010, given the following:

class Base { };
class Derived : public Base { };

template<class T> void foo(T& t);  // A
void foo(Base& base);              // B

Derived d;
foo(d);                            // calls A
foo(static_cast<Base&>(d));        // calls B

我想在上面叫"B".我可以通过强制转换为Base来实现这一点，但是为什么这样做是必需的?

I would like "B" to be called above. I can achieve this with a cast to Base, but why is this necessary?

我希望为不是从Base派生的所有类型(内置类型等)调用模板函数，但是我希望为从Base派生的类型调用非模板重载，而没有要求客户明确投放.我还尝试使重载成为模板的特殊化，但是在这种情况下会发生相同的行为.获得我要找的东西的惯用方式是什么?

I want the template function to be called for all types not derived from Base (built-in types, etc.), but I want the non-template overload to be called for types derived from Base, without requiring the client to explicitly cast. I also tried making the overload a specialization of the template, but the same behavior occurs in that case. What is the idiomatic way to get what I'm looking for?

推荐答案

在所有条件都相同的情况下，非模板函数优于函数模板.但是，在您的情况下，所有条件都不相等:(A)与T = Derived完全匹配，但是(B)需要参数的派生到基数转换.

All things being equal, nontemplate functions are preferred over function templates. However, in your scenario, all things are not equal: (A) is an exact match with T = Derived, but (B) requires a derived-to-base conversion of the argument.

您可以通过使用SFINAE(替代失败不是错误)来解决特定情况(如这种情况)的情况，以防止使用从Base派生的类型实例化(A):

You can work around this for specific cases (like this one) by using SFINAE (substitution failure is not an error) to prevent (A) from being instantiated with a type that is derived from Base:

#include <type_traits>
#include <utility>

template <typename T>
typename std::enable_if<
    !std::is_base_of<Base, T>::value
>::type foo(T& x)
{
}

void foo(Base& x)
{
}

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