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

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有一个基类？看来它不能,基于以下测试.请在最后查看我发表问题的评论.

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));
    }
};

解决方法:

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

这并不像你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这种强制转换,这导致后者在重载解析期间更好地匹配.

要强制模板功能与Base和Derived不匹配,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

标签：c,templates,pointers,inheritance,template-function
来源： https://codeday.me/bug/20191009/1875933.html