要求四个线程 抢占线程 输出1~100
要求用到lock unlock
累似 如下C++代码思路(下面的unlock有点儿问题)
#include
#include
#include//互斥锁
using namespace std;//调用命名空间std内定义的所有标识符
mutex mutex_number;
const int MAXNUM = 100;
int number = 0;
void outNumber(int index)
{
while (1) {
mutex_number.lock();
if (number >= MAXNUM) {
mutex_number.unlock();
break;
}
number++;
cout << "mythread_ " << index + 1 << ": " << number << endl; // 输出
mutex_number.unlock();
}
}//抢占线程
int main()
{
//声明四个线程
thread mythread[4];
for (int i = 0; i < 4; i++)
{
mythread[i] = thread(outNumber,i);//第一个参数为函数名,第二个为线程参数
}
// 阻塞线程,等待加入
for (int i = 0; i < 4; i++)
mythread[i].join();
return 0;
}
这样?有帮助麻烦点个采纳【本回答右上角】,谢谢~~
using System;
using System.Threading;
namespace ConsoleApplication1
{
class Program
{
int num = 0, maxnum = 100;
void outNumber(object index)
{
while (true)
{
Monitor.Enter(this);
if (num >= maxnum) break;
num++;
Console.WriteLine("线程" + index + ":" + num);
Monitor.Exit(this);
}
}
static void Main(string[] args)
{
var p = new Program();
var pools = new Thread[4];
for (var i = 0; i < 4; i++)
{
pools[i] = new Thread(p.outNumber);
}
for (var i = 0; i < 4; i++)
{
pools[i].Start(i);
}
Console.ReadKey();
}
}
}