Question

参考http://www.yolinux.com/TUTORIALS/LinuxTutorialPosixThreads.html#SCHEDULING

我试图在C ++中创建两个线程并尝试将字符串作为参数传递给Thread Start Routine。根据定义，Thread Start Routine参数只能是(void *)类型：

int pthread_create(pthread_t * thread, 
                       const pthread_attr_t * attr,
                       void * (*start_routine)(void *), 
                       void *arg);

但我得到以下错误：

$ make
g++ -g -Wall Trial.cpp -o Trial
Trial.cpp: In function `int main()':
Trial.cpp:22: error: cannot convert `message1' from type `std::string' to type `void*'
Trial.cpp:23: error: cannot convert `message2' from type `std::string' to type `void*'
Makefile:2: recipe for target `Trial' failed
make: *** [Trial] Error 1

代码是

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



using namespace std;

void *print_message_function( void *ptr );


int main()
{
    pthread_t thread1, thread2;

    string message1 = "Thread 1";
    string message2 = "Thread 2";

    int  iret1, iret2;


     iret1 = pthread_create( &thread1, NULL, print_message_function, (void*) message1);
     iret2 = pthread_create( &thread2, NULL, print_message_function, (void*) message2);



     pthread_join( thread1, NULL);
     pthread_join( thread2, NULL);



     cout << "Thread 1 returns: " <<  iret1 << endl;
     cout << "Thread 2 returns: " << iret2 << endl;


    return 0;
    }
void *print_message_function( void *ptr )

{
     cout << endl <<  ptr << endl;
     //return 0;

}

有没有办法将string作为(void *)参数传递？或者只有C style strings可以用作多线程参数 - 如链接中的参考代码所示。

Answer 1

参数需要是指针，并且您尝试将对象传递给它。

您有两个选择，要么将指针传递给std::string对象，要么将指针传递给基础字符串。我推荐第一个：

 iret1 = pthread_create(&thread1, NULL, print_message_function, &message1);

然后你必须修改线程函数，否则它将打印指针而不是它指向的字符串：

void* print_message_function(void* ptr)
{
    std::string str = *reinterpret_cast<std::string*>(ptr);

    std::cout << str << std::endl;

    return nullptr;
}

除非要求使用POSIX线程，否则我宁愿推荐C++ standard library中的线程功能：

#include <iostream>
#include <string>
#include <thread>

void print_message_function(const std::string& msg);

int main()
{
    std::string message1 = "Thread 1";
    std::string message2 = "Thread 2";

    std::thread thread1(print_message_function, std::cref(message1));
    std::thread thread2(print_message_function, std::cref(message2));

    thread1.join();
    thread2.join();
}

void print_message_function(const std:string& msg)
{
    std::cout << msg << std::endl;
}

Answer 2

您可以通过将所有参数包装到简单结构中来将任何参数传递给pthread_create方法。例如：

struct ThreadParams {
    std::vector<int> ints;
    std::string clientName;
    // more params
};

你需要做的就是在调用CreateThread函数之前初始化这个结构，然后传递一个指针：

ThreadParams * params = new ThreadParams();
params.setParameters();
pthread_create(..., params);

void* print_message_function(void* arg)
    ThreadParams * params = reinterpret_cast<ThreadParams *>(arg);
    // delete after usage;
    delete params;
}

Answer 3

调用pthread_create时需要获取字符串对象的地址（内存位置）：

iret1 = pthread_create( &thread1, NULL, print_message_function, (void*)&message1);

当你打印它时你需要将你的内存void *转换回字符串指针（你不能写一个void *到一个流，流不会知道如何处理它），取消引用它：

cout << endl <<  *(string*)ptr << endl;

这应该可行，但为了完全正确，你应该看一下在向*和从*转换时使用reinterpret_cast。

将字符串作为Thread启动例程参数传递：C ++

3 个答案: