为什么在C ++中调用此虚函数会导致基类出现“缺少符号”错误?

时间:2015-04-11 15:52:35

标签: c++ templates linker-errors

我有一个虚拟类,由以下内容定义:

  template<typename TId, typename TValue>
  class Resource {
    private:

      /// Next item in the chain
      unique_ptr<Resource<TId, TValue>> data;

    protected:

      /// The id this value
      TId id;

      /// The value on this value
      TValue value;

    public:

      /// Create a new instance and assign the values to it
      Resource(TId id, TValue value) {
        this->id = id;
        this->value = value;
      }

      /// Destructor
      virtual ~Resource() {
      }

      /// Safely clone a new instance of this
      virtual Resource<TId, TValue> *Clone();

      ... other stuff removed for brevity ...

      /// Create a new resource chain with a custom filter
      Option<Resource<TId, TValue>*> Filter(std::function< bool(Resource<TId, TValue>*) > filter) {
        auto iter = this->Iter<Resource<TId, TValue>>(filter);
        if (iter.Any()) {
          auto index = 0;
          auto root = (*iter.First())->Clone();
          iter.Each([&] (Resource<TId, TValue>* r) {
            if (index != 0) {
              root->Push(r->Clone());
            }
            ++index;
          });
          return Some<Resource<TId, TValue>*>(root);
        }
        return None<Resource<TId, TValue>*>();
      }
  };

我在测试中通过以下方式实现了这一点:

enum RType {
  ThingOne = 1,
  ThingTwo = 2
};

class RValue : public Resource<RType, i32> {
  public:
    RValue(int value) : Resource(ThingOne, value) {
    }
    ~RValue() {
    }
    Resource<RType, i32>* Clone() {
      return new RValue(this->value);
    }
};

NB。请注意克隆的使用:

root->Push(r->Clone());

然而,在编译时我得到:

[ 88%] Building CXX object CMakeFiles/test-resource.dir/tests/test-resource.cpp.o
Linking CXX executable tests/test-resource
Undefined symbols for architecture x86_64:
  "npp::Resource<RType, int>::Clone()", referenced from:
      vtable for npp::Resource<RType, int> in test-resource.cpp.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
make[2]: *** [tests/test-resource] Error 1
make[1]: *** [CMakeFiles/test-resource.dir/all] Error 2

编译器是clang,平台是OSX。

发生了什么?

为什么在调用函数时,派生类中的实现不会自动解析?

是因为这是一个模板化的方法吗?

完整的代码可以在这个要点中找到,供参考:https://gist.github.com/shadowmint/d49650668e9a74c324a1

1 个答案:

答案 0 :(得分:5)

如果您不想实现基类的虚拟成员函数,则必须将其声明为纯虚拟:

  //                                     vvv-- here
  virtual Resource<TId, TValue> *Clone() = 0;

否则链接器将搜索您声明但在生成基类的虚函数表时未实现的函数(您可以在错误消息中看到)。

请注意,在其中声明纯虚拟成员函数将使类成为抽象。