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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，它似乎不能。有关问题发生的位置，请参阅我最后的评论。

我怎样才能让它工作？

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));
    }
};
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1 回答 1

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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 *to的转换；Base *但是模板版本不需要这种转换，这导致后者在重载解析期间更好地匹配。

要强制模板函数不匹配两者Base，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
于 2012-08-10T23:46:29.867 回答