如何通过引用pthread启动例程传递顺序计数器? [英] How to pass a sequential counter by reference to pthread start routine?

问题描述

下面是我的C代码，用于打印一个递增的全局计数器，每个线程一个增量.

Below is my C code to print an increasing global counter, one increment per thread.

#include <stdio.h>
#include <pthread.h>

static pthread_mutex_t pt_lock = PTHREAD_MUTEX_INITIALIZER;
int count = 0;

int *printnum(int *num) {
    pthread_mutex_lock(&pt_lock);
    printf("thread:%d ", *num);
    pthread_mutex_unlock(&pt_lock);
    return NULL;
}

int main() {
    int i, *ret;
    pthread_t pta[10];
    for(i = 0; i < 10; i++) {
        pthread_mutex_lock(&pt_lock);
        count++;
        pthread_mutex_unlock(&pt_lock);
        pthread_create(&pta[i], NULL, (void *(*)(void *))printnum, &count);
    }
    for(i = 0; i < 10; i++) {
        pthread_join(pta[i], (void **)&ret);
    }
}

我希望每个线程打印全局计数器的一个增量，但它们错过增量，有时会从两个线程访问全局计数器的相同值.如何使线程顺序访问全局计数器?

I want each thread to print one increment of the global counter but they miss increments and sometimes access same values of global counter from two threads. How can I make threads access the global counter sequentially?

示例输出:

线程:2 线程数:3 线程数:5 线程数:6 线程数:7 线程数:7 线程数:8 线程数:9 线数:10 线程:10

thread:2 thread:3 thread:5 thread:6 thread:7 thread:7 thread:8 thread:9 thread:10 thread:10

修改

蓝月亮的答案解决了这个问题. MartinJames的评论中提供了替代方法.

Blue Moon's answer solves this question. Alternative approach is available in MartinJames'es comment.

推荐答案

一种简单但无用的方法是确保thread1打印1，thread2打印2，以此类推将 join 立即线程:

A simple-but-useless approach is to ensure thread1 prints 1, thread2 prints 2 and so on is to put join the thread immmediately:

pthread_create(&pta[i], NULL, printnum, &count);
pthread_join(pta[i], (void **)&ret);

但这完全破坏了多线程的目的，因为一次只能有一个进程.

But this totally defeats the purpose of multi-threading because only one can make any progress at a time.

请注意，我删除了多余的强制类型转换，线程函数也接受了void *参数.

Note that I removed the superfluous casts and also the thread function takes a void * argument.

一种更明智的方法是按值传递循环计数器i，以便每个线程将打印不同的值，并且您会看到线程正在运行，即编号1-10可以按任何顺序打印，并且每个线程也会打印一个唯一值.

A saner approach would be to pass the loop counter i by value so that each thread would print different value and you would see threading in action i.e. the numbers 1-10 could be printed in any order and also each thread would print a unique value.

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