c++ - 对虚拟表、构造函数和析构函数的 undefined reference

Question

我在处理我正在处理的一些许可多任务代码的构造函数和析构函数时遇到了问题。我收到的错误是

task.o: In function `$_4Task':
/home/luke/project/task.cc(.text+0x57): undefined reference to `Task virtual table'
task.o: In function `_4TaskUi':
/home/luke/project/task.cc(.text+0x5): undefined reference to `Task virtual table'

从目标文件来看，_4Task 和 _4TaskUi 的取消引用名称分别是我的任务析构函数和构造函数。我似乎无法弄清楚为什么我的编译器告诉我我没有引用虚拟表。任何帮助将不胜感激，我在 ubuntu 12.10 上使用 GCC

下面是我的头文件

#ifndef TASK
#define TASK

#include "stddef.h"
#include "nocopy.h"

class Task: private NoCopy
{
    void** stack;           //Base of stack
    void** sp;              //saved sp when not running

    static Task* current;               //Point to running task
    static void start();                //calls task::main
    static void dispatch(Task* t);      //switch context to task t
    virtual void main();
    friend class TaskList;

protected:
    void fork();            //start task

public:
    Task(size_t stackSize);
    virtual ~Task();
    static Task* const getCurrent()
    {
        return current;
    }
    static void yield();
};

 #endif

下面是我的cpp文件

#include "task.h"

Task initialTask(0);                //initial code before 
                             // set up stack in crt0.s
Task* Task::current = &initialTask;         //note first running task
TaskList readyList;
Task::Task(size_t stackSize)
:stack(new (void*)[stackSize/sizeof(void*)]),
 sp(&stack[stackSize/sizeof(void*)])
{                               //set up task stack
    if(stackSize)
    {
        *--sp = 0;      //for debugger
        *--sp = 0;      //terminate frame chain
        *--sp = &start; //point to first code
    }
}

Task::~Task()
{
    delete[](stack);
}
//Contec Switching
register void** spreg asm("s");     //can refer to hc12 SP as spreg
void Task::dispatch(Task* task)
{
    current -> sp = spreg;
    current = task;
    spreg = current -> sp;
}//Dispatch is called by one task but is returned by another

void Task::fork()       //call in ctor of all task
{
    TaskList::Node node(&readyList);    //Make caller go from "running"->"ready"
    dispatch(this);                 //Start new task
}

void Task::start()
{
    current -> main();
    TaskList forever;       //Wait Task
    forever.enqueue();      //Wait forever  
}

编辑 我想通了，因为每个人都说 main 没有在 task.h 或 task.cc 中定义，main 取决于任务，所以 main 的默认定义是

virtual void main(){};  

Answer 1

您需要为您的类中的所有虚拟 函数提供定义。只有纯虚函数可以在没有定义的情况下存在。您显示的代码没有函数 Task::main() 的任何定义。你需要定义它。
再三考虑，我会将该函数命名为更合适的名称。

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好读:

What does it mean that the "virtual table" is an unresolved external?