使用继承时未解析的引用

Question

如何将模板定义转换为单独的头文件？

我的代码在包含在单个main.cpp中时编译.

maip.cpp

#include <windows.h>
#include <tchar.h>

template<class T1, class T2>
class Base {
public:
    virtual ~Base() {}
    virtual T1 f1();
    virtual T2 f2();
};

template<class T1, class T2>
T1 Base<T1, T2>::f1() {return T1();}
template<class T1, class T2>
T2 Base<T1, T2>::f2() {return T2();}

class Derived : public Base<int, int> {
public:
    virtual ~Derived() {}
};

int _tmain(int argc, _TCHAR* argv[])
{
    int i;
    Derived d;
    i = d.f1();
    return 0;
}

但是,当我分解时,我得到了未解析的外部符号:

main.cpp中

#include <windows.h>
#include <tchar.h>

#include "Base.h"
#include "Derived.h"

int _tmain(int argc, _TCHAR* argv[])
{
    int i;
    Derived d;
    i = d.f1();
    return 0;
}

Base.h

#pragma once

template<class T1, class T2>
class Base {
public:
    Base();
    virtual ~Base() {}
    virtual T1 f1();
    virtual T2 f2();
};

template<class T1, class T2>
T1 Base<T1, T2>::f1() {return T1();}
template<class T1, class T2>
T2 Base<T1, T2>::f2() {return T2();}

Derived.h

#pragma once

class Derived : public Base<int, int> {
public:
    virtual ~Derived() {}
};

这导致:

错误1错误LNK2019:未解析的外部符号"public:__thiscall Base :: Base(void)"(?? 0？$ Base @ HH @@ QAE @ XZ)在函数"public:__thiscall Derived :: Derived(void)"中引用(?? 0Derived @@ @ QAE XZ)

Answer 1

您没有为Base构造函数提供实现.尝试添加

template<class T1, class T2>
Base<T1, T2>::Base () { }

或者,您可以从第二个代码段中删除其声明.(它也不在您的第一个代码示例中).

您的链接器基本上表示Derived::Derived ()尝试调用Base::Base (),您明确声明了,但他无法找到它的实现.