c++ - 为什么模板函数不能将指向派生类的指针解析为指向基类的指针

Question

编译器是否无法在编译时获取指向派生类的指针并知道它有一个基类？根据以下测试，似乎不能。请参阅我在末尾的评论以了解问题发生的位置。

我怎样才能让它发挥作用？

std::string nonSpecStr = "non specialized func";
std::string const specStr = "specialized func";
std::string const nonTemplateStr = "non template func";

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


template <typename T> std::string func(T * i_obj)
{ return nonSpecStr; }

template <> std::string func<Base>(Base * i_obj)
{ return specStr; }

std::string func(Base * i_obj)
{ return nonTemplateStr; }

class TemplateFunctionResolutionTest
{
public:
    void run()
    {
        // Function resolution order
        // 1. non-template functions
        // 2. specialized template functions
        // 3. template functions
        Base * base = new Base;
        assert(nonTemplateStr == func(base));

        Base * derived = new Derived;
        assert(nonTemplateStr == func(derived));

        OtherClass * otherClass = new OtherClass;
        assert(nonSpecStr == func(otherClass));


        // Why doesn't this resolve to the non-template function?
        Derived * derivedD = new Derived;
        assert(nonSpecStr == func(derivedD));
    }
};

Answer 1

Derived * derivedD = new Derived;
assert(nonSpecStr == func(derivedD));

这不会像您期望的那样解析为非模板函数，因为要做到这一点，必须执行从 Derived * 到 Base * 的转换；但模板版本不需要此转换，这导致后者在重载解析期间具有更好的匹配性。

要强制模板函数不匹配 Base 和 Derived，您可以使用 SFINAE 拒绝这两种类型。

#include <string>
#include <iostream>
#include <type_traits>
#include <memory>

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

template <typename T> 
typename std::enable_if<
    !std::is_base_of<Base,T>::value,std::string
  >::type
  func(T *)
{ return "template function"; }

std::string func(Base *)
{ return "non template function"; }

int main()
{
  std::unique_ptr<Base> p1( new Base );
  std::cout << func(p1.get()) << std::endl;

  std::unique_ptr<Derived> p2( new Derived );
  std::cout << func(p2.get()) << std::endl;

  std::unique_ptr<Base> p3( new Derived );
  std::cout << func(p3.get()) << std::endl;

  std::unique_ptr<OtherClass> p4( new OtherClass );
  std::cout << func(p4.get()) << std::endl;
}

输出:

non template function
non template function
non template function
template function