c++ - 为何模板函数无法将指向派生类的指针解析为指向基类的指针

编译器在编译时是否无法获取指向派生类的指针并知道它具有基类？ 根据以下测试，似乎不能。有关问题发生的位置，请参阅我的评论结尾。

如何使它工作？

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));
    }
};

最佳答案

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

这不能像您期望的那样解析为非模板函数，因为这样做必须执行从Derived *到Base *的转换。但是模板版本不需要此强制类型转换，从而导致后者在重载解析期间更好地匹配。

要强制模板函数不同时匹配Base和Derived，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

关于c++ - 为何模板函数无法将指向派生类的指针解析为指向基类的指针，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/11910558/