对具有实现的函数的未定义引用

undefined reference to functions that have implementation

本文关键字:函数 引用 未定义 实现      更新时间:2023-10-16

我正在使用带有标头和 cpp 文件的线程类。 当我把它们都放在空测试文件时,它写道:

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

我只是试图将Thread.h和Thread组合在一起.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 exepctionn";
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;
}

您在这里面临的问题不是构建问题,而是链接问题。当构建thread.o时,编译器知道pthread_create存在,并在某处定义,因为在pthread.h头中声明。

如果您使用nm查看thread.o中使用的符号,您将看到类似以下内容的内容:

U _pthread_create
U _pthread_detach
...

这告诉您thread.o引用多个Undefined 符号,包括pthread_create。换句话说,此时pthread_create的机器代码是未知的。在您需要将目标文件链接到可执行文件(链接器的角色(之前,这完全没问题。

在此阶段,必须告知链接器在哪里可以找到这些未定义的符号,可能来自另一个对象文件或静态/共享库。对于pthread,符号在libpthread中定义,你可以在系统目录中找到libpthread.a。您可以通过添加-lpthread来告诉g++链接此库(请注意,使用-l时省略了libpthread中的lib(:

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

通常,如果使用静态库中引用的符号lib${LIBNAME}.a${LIBDIR}目录中可用的符号,则可以告诉链接器将其用于:

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