标签：lin img argv 问题： start 同步 work 函数 情况
串行解决方案示例代码：
#include <QCoreApplication>
#include <QThread>
#include <QDebug>
#include <QObject>
/*
sum(n)=> 1 + 2 + 3 + ..... + n
sum(1000)= ?
[1,1000] = [1,300] + [301,600] + [601, 1000]
*/
class Calculator : public QObject
{
protected:
int m_begin;
int m_end;
int m_result;
void run()
{
qDebug() << objectName() << " run begin";
for(int i=m_begin; i<=m_end; i++)
{
m_result += i;
}
qDebug() << objectName() << " run end";
}
public:
Calculator(int begin,int end)
{
m_begin = begin;
m_end = end;
m_result = 0;
}
void work()
{
run();
}
int result()
{
return m_result;
}
};
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
qDebug() << "main begin()" ;
int sum = 0;
Calculator c1(1,300);
Calculator c2(301,600);
Calculator c3(601,1000);
c1.setObjectName("c1");
c2.setObjectName("c2");
c3.setObjectName("c3");
c1.work();
c2.work();
c3.work();
sum = c1.result() + c2.result() + c3.result();
qDebug() << "sum = " << sum;
qDebug() << "main end()";
return a.exec();
}
从打印结果看，c1先运行，c2再运行，最后c3再运行。这种串行解决方案完全可以解决求和问题，但是太低效了。
编程实验：求和新解法——并行方案的代码示例：
#include <QCoreApplication>
#include <QThread>
#include <QDebug>
/*
sum(n)=> 1 + 2 + 3 + ..... + n
sum(1000)= ?
[1,1000] = [1,300] + [301,600] + [601, 1000]
*/
class Calculator : public QThread
{
protected:
int m_begin;
int m_end;
int m_result;
void run()
{
qDebug() << objectName() << " run begin";
for(int i=m_begin; i<=m_end; i++)
{
m_result += i;
msleep(10); //模拟在实际工程中可能产生耗时的行为
}
qDebug() << objectName() << " run end";
}
public:
Calculator(int begin,int end)
{
m_begin = begin;
m_end = end;
m_result = 0;
}
int result()
{
return m_result;
}
};
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
qDebug() << "main begin()" ;
int sum = 0;
Calculator c1(1,300);
Calculator c2(301,600);
Calculator c3(601,1000);
c1.setObjectName("c1");
c2.setObjectName("c2");
c3.setObjectName("c3");
c1.start();
c2.start();
c3.start();
sum = c1.result() + c2.result() + c3.result();
qDebug() << "sum = " << sum;
qDebug() << "main end()";
return a.exec();
}
从打印结果看，主线程先于子线程结束，c1，c2,c3这三个进程确实可以并行执行，执行速度也比串行解决方案快，但是执行结果却是错误的。为什么？
问题：线程间总是完全独立毫无依赖吗？
结论：在特殊情况下，多线程的执行在时序上存在依赖。
编程实验，并行计算初探
#include <QCoreApplication>
#include <QThread>
#include <QDebug>
/*
sum(n)=> 1 + 2 + 3 + ..... + n
sum(1000)= ?
[1,1000] = [1,300] + [301,600] + [601, 1000]
*/
class Calculator : public QThread
{
protected:
int m_begin;
int m_end;
int m_result;
void run()
{
qDebug() << objectName() << " run begin";
for(int i=m_begin; i<=m_end; i++)
{
m_result += i;
msleep(10); //模拟在实际工程中可能产生耗时的行为
}
qDebug() << objectName() << " run end";
}
public:
Calculator(int begin,int end)
{
m_begin = begin;
m_end = end;
m_result = 0;
}
int result()
{
return m_result;
}
};
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
qDebug() << "main begin()" ;
int sum = 0;
Calculator c1(1,300);
Calculator c2(301,600);
Calculator c3(601,1000);
c1.setObjectName("c1");
c2.setObjectName("c2");
c3.setObjectName("c3");
c1.start();
c2.start();
c3.start();
/*在这个地方主线程需要等待3个子线程执行结束，才能再执行*/
c1.wait();
c2.wait();
c3.wait();
sum = c1.result() + c2.result() + c3.result();
qDebug() << "sum = " << sum;
qDebug() << "main end()";
return a.exec();
}
效率提高了，运行结果也完全正确。
看一下wait函数的介绍，在linux编程中，这个函数类似于pthread_join()
标签：lin img argv 问题： start 同步 work 函数 情况
原文地址：https://www.cnblogs.com/-glb/p/13379912.html
