对具有实现功能的未定义引用

时间:2019-10-12 13:57:28

标签: c++

我正在使用带有标头和cpp文件的线程类。 当我把它们都放到空的测试文件中时,它写着:

g++ -g -pedantic -ansi -Wall -Werror -std=c++03 -I../include  -c -o test.o test.cpp
g++ -g  test.o thread.o   -o test
thread.o: In function `Thread::~Thread()':
/home/tomer/work/mt/hash2/cpp/thread.cpp:15: undefined reference to `pthread_detach'
thread.o: In function `Thread::start()':
/home/tomer/work/mt/hash2/cpp/thread.cpp:40: undefined reference to `pthread_create'
thread.o: In function `Thread::join()':
/home/tomer/work/mt/hash2/cpp/thread.cpp:49: undefined reference to `pthread_join'
thread.o: In function `Thread::cancel()':
/home/tomer/work/mt/hash2/cpp/thread.cpp:58: undefined reference to `pthread_cancel'
thread.o: In function `Thread::detach()':
/home/tomer/work/mt/hash2/cpp/thread.cpp:66: undefined reference to `pthread_detach'
collect2: error: ld returned 1 exit status
<builtin>: recipe for target 'test' failed
make: *** [test] Error 1

我只是尝试编译Thread.h和Thread.cpp

//Thread.h looks like this:
#ifndef THREAD_H
#define THREAD_H

#include <cstddef>
#include <pthread.h>
#include <string>

class Thread
{
public:
    Thread(size_t a_userID = 0);
    virtual ~Thread();

    bool start();
    void join();
    void cancel();
    void detach();


private:
    static void* threadMainFunction(void *);
    virtual void run() = 0;
    bool isAlive(std::string a_msg);

private:
    bool m_joinable;

protected:
    pthread_t m_threadID;
    size_t m_userID;
};

#endif

//Thread.cpp looks like this:
#include <exception>
#include "Thread.h"
#include <iostream>
Thread::Thread(size_t a_userID)
: m_joinable(true)
, m_threadID(0)
, m_userID(a_userID)
{
}

Thread::~Thread()
{
    if(m_joinable)
    {
        pthread_detach(m_threadID);
    }
}

void* Thread::threadMainFunction(void *a_thread)
{
    Thread* thread = reinterpret_cast<Thread*>(a_thread);
    try
    {
        thread->run();
    }
    catch(const std::exception& e)
    {
        std::cout<<"what exepction\n";
        std::cerr << e.what() << '\n';
    }
    catch(...)
    {
        throw;
    }
    return 0;
}

bool Thread::start()
{
    int r = pthread_create(&m_threadID, 0, threadMainFunction, this);
    return r == 0;
}

void Thread::join()
{
    if(isAlive("Thread::join on thread not started"))
    {
        void *status;
        pthread_join(m_threadID, &status);
        m_joinable = false;
    }
}

void Thread::cancel()
{
    if(isAlive("Thread::cancel on thread not started"))
    {
        pthread_cancel(m_threadID);
    }
}

void Thread::detach()
{
    if(isAlive("Thread::detach on thread not started"))
    {
        pthread_detach(m_threadID);
    }
}

bool Thread::isAlive(std::string a_msg)
{
    if(m_threadID == 0)
    {
        throw(std::runtime_error(a_msg));
        return false;
    }
    return true;
}

1 个答案:

答案 0 :(得分:1)

您在此处面临的问题不是构建问题,而是链接问题。在构建thread.o时,编译器知道pthread_create存在,并且在pthread.h标头中声明了,因此定义在某处。

如果您使用nm来查看thread.o中使用的符号,则会看到类似的内容:

U _pthread_create
U _pthread_detach
...

这告诉您thread.o引用了多个 U 未定义符号,包括pthread_create。换句话说,此时pthread_create的机器代码是未知的。这非常好,直到您需要将目标文件链接到可执行文件中,该文件是链接器的角色。

在此阶段,您必须告诉链接器在哪里可以找到这些未定义的符号,可能是从另一个目标文件或静态/共享库中找到的。对于pthread,符号在libpthread中定义,您可能会在系统目录中找到libpthread.a。您可以通过添加g++来告诉-lpthread链接此库(请注意,使用lib时,libpthread中的-l被省略了)

g++ -g test.o thread.o -o test -lpthread

通常,如果您使用lib${LIBNAME}.a目录中可用的静态库${LIBDIR}中引用的符号,则可以告诉链接程序将其用于:

g++ -g *.o -L$LIBDIR -l${LIBNAME}