Question

所以我开始编写一个解析然后处理大量文本的程序。我已经设置了一个包含作为boost线程运行的方法的类。截至目前，这些线程中的每一个都只是打印一些文本语句然后返回。代码编译并运行没有任何错误。但是，文本打印不一致。当线程并行运行时，这是预期的，所以我尝试使用互斥锁来协调输出的使用。但是，由于输出仍然不一致，我显然做错了。要添加到此，一些输出打印两次，我无法解释为无法正确编码互斥锁。以下是我的代码：

/* 
 * File:   ThreadParser.h
 * Author: Aaron Springut
 *
 * Created on Feburary 2, 2012, 5:13 PM
 */

#ifndef THREADPARSER_H
#define THREADPARSER_H

#include <string.h>
#include <iostream>
#include <boost/thread/thread.hpp>
#include <boost/thread/mutex.hpp>
#include <boost/bind.hpp>
#include "FileWordCounter.h"

class ThreadParser {

    public:

    ThreadParser();
    ThreadParser(std::string fileName);



    private:

    //mutex for cout
    boost::mutex coutMut;  

    std::string dataFile;
    FileWordCounter fCounter;

    //threads
    void parseFile();
    void processSearches();



};

#endif     


/* 
 * File:   ThreadParser.cpp
 * Author: Aaron Springut
 *
 * Created on Feburary 2, 2012, 5:13 PM
 */


 #include "ThreadParser.h"

 using namespace std;


 ThreadParser::ThreadParser(){

    double buyNum = 0;
    buyNum = buyNum * 100;
    cout << "Percentage of people buying: "<< buyNum <<"%"<<endl;

 }

 ThreadParser::ThreadParser(string fileName){

    dataFile = fileName;
    double buyNum = 0;

    //create the mutex and aquire a lock on it for this thread

    boost::mutex::scoped_lock(coutMut);

    boost::thread parseThread(boost::bind(&ThreadParser::parseFile, this));
    boost::thread processSearches(boost::bind(&ThreadParser::processSearches,this));

    buyNum = buyNum * 100;
    cout << "Percentage of people buying: "<< buyNum <<"%"<<endl;


 }


 void ThreadParser::parseFile(){

    boost::mutex::scoped_lock(coutMut);
    cout << "parseFileThreadLaunch"<<endl;
    return;

 }


 void ThreadParser::processSearches(){

    boost::mutex::scoped_lock(coutMut);
    cout << "processSearchesLaunch"<<endl;
    return;

 }

这里出现问题的一个例子是程序运行的两个输出：

Percentage of people buying: parseFileThreadLaunch
processSearchesLaunch
0%

好的，cout不是线程安全的，我对互斥锁做错了。

Percentage of people buying: parseFileThreadLaunch
0%
processSearchesLaunch
processSearchesLaunch

这令人困惑，最后一行怎么打印两次？这是cout不是线程安全的结果吗？或者，我错过了大局的一部分。

编辑：在主函数中调用类是这样的：

string fileName("AOLData.txt");
cout << fileName<< endl;

ThreadParser tp(fileName);

Answer 1

boost::mutex::scoped_lock(coutMut);

这不符合你的想法。这会创建一个临时的，立即销毁，释放锁。就像int(3);。

一样

你想：

boost::mutex::scoped_lock sl(moutMut);

这将创建一个对象sl，该对象持有锁，直到它超出范围。

使用boost线程库进行打印

1 个答案: