信号量不工作 C++

Question

我是信号量的新手，正在尝试编写一个非常基本的示例以更好地学习它。我的代码包含三个数组：

b1[5]={1;2;3;4;5}
b2[5]={6;7;8;9;10}
x[10]

我有一个函数接受 b1 或 b2 并更新 x 使其如下：

x={0;1;2;3;4;5;6;7;8;9}

我的代码如下：

#include <sys/types.h>  /* Primitive System Data Types */ 
#include <errno.h>      /* Errors                      */
#include <stdio.h>      /* Input/Output                */
#include <stdlib.h>     /* General Utilities           */
#include <thread>       /* C++11 Threads               */
#include <string.h>     /* String handling             */
#include <semaphore.h>  /* Semaphore                   */
#include <Windows.h>    /* windows                     */
using namespace std;

/* prototype for thread routine */
void handler(int x[], int b[], int start, int id);

/* global vars */
sem_t mutex;
vector<std::thread> threadList;

int main()
{
int i[2];
int x[10];
int b1[5];
int b2[5];
sem_init(&mutex, 0, 1);      /* initialize mutex to 1 - binary semaphore     */
/* second param = 0 - semaphore is local */
i[0] = 0;
i[1] = 1;

    // Update the matrix
    for (int j = 0; j < 5; j++)
    {
        b1[j] = j;
        b2[j] = j+5;
    }



        threadList.push_back(std::thread(handler, x, b1, 0, 0));
        threadList.push_back(std::thread(handler, x, b2, 5, 1));

        // wait for all threads to finish
        for (auto& threadID : threadList){
            threadID.join();
        }

sem_destroy(&mutex); /* destroy semaphore */

/* exit */
return 0;
} /* main() */

void handler(int x[], int b[], int start, int id)
{

    for (int j = start; j < 5; j++)
    {
        x[j] = b[j];
     }
    printf("Thread %d: Waiting to print results...\n", id);
    sem_wait(&mutex);       /* down semaphore */
    /* START CRITICAL REGION */
    for (int j = start; j < 5; j++)
    {
        printf("x[%d] = %d\n", j , x[j]);
    }
    /* END CRITICAL REGION */
    sem_post(&mutex);       /* up semaphore */
}

代码的输出如下：

Thread 0: Waiting to print results...
x[0] = 0
x[1] = 1
x[2] = 2
x[3] = 3
x[4] = 4
Thread 1: Waiting to print results...

然而，我期待的是以下内容

Thread 0: Waiting to print results...
x[0] = 0
x[1] = 1
x[2] = 2
x[3] = 3
x[4] = 4
Thread 1: Waiting to print results...
x[5] = 5
x[6] = 6
x[7] = 7
x[8] = 8
x[9] = 9

知道为什么第二个线程没有进入代码的打印部分吗？

Answer 1

您对 'handler' 的第二次调用开始=5。

循环

for (int j = start; j< 5; j++)

两个地方都没有执行，因为j >= 5.

信号量不工作 C++

Semaphore not working C++

multithreading

semaphore

c++11