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
#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)
1 回答
您没有为
Base
的构造函数提供实现 . 尝试添加或者,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) .
你的链接器基本上说
Derived::Derived ()
试图调用你明确声明的Base::Base ()
,但他找不到它的实现 .