如何产卵n个线程？ [英] How to spawn n threads?

问题描述

我想写一个多线程程序，基于命令行输入线程数，所以我不能硬code pre-声明线程。这是做这件事的一个有效的方法是什么？

  INT线程= 5; //（动态的，不是硬codeD）
INT I = 0;
*的pthread_t线程=的malloc（sizeof的（的pthread_t）*线程）;对于（i = 0; I＆LT;螺纹;我++）{
    的pthread_t foobar的;
    线程[I] = foobar的; //这将导致冲突？
}对于（i = 0; I＆LT;螺纹;我++）{    INT RET =在pthread_create（安培;螺纹[I]，NULL，（void *的）及foobar_function，NULL）;    如果（保留！= 0）{
        的printf（创建pthread的错误。\\ n！）;
        出口（1）;
    }
}
 

---

下面是修改我的结果以下建议。似乎工作就好了。

  INT线程= 5;
INT I;*的pthread_t线程=的malloc（sizeof的（的pthread_t）*线程）;对于（i = 0; I＆LT;螺纹;我++）{    INT RET = pthread_create的（安培;螺纹[Ⅰ]，NULL，＆放大器; foobar_function，NULL）;    如果（保留！= 0）{
        的printf（创建pthread的错误。\\ n！）;
        出口（1）;
    }
    //在pthread_join（线程[I]，NULL）; //实际上并不希望这样在这里:)
}睡眠（1）; //主（）你的线程之前做可能会结束，
免费（螺纹）; //所以我们睡用于说明目的
 

解决方案

什么是在第一个周期？它是否设置数组元素为未初始化价值？

所以，我认为这就是你所需要的：

  INT线程= 5，I = 0，RET = -1;*的pthread_t线程=的malloc（sizeof的（的pthread_t）*线程）;对于（i = 0; I＆LT;螺纹;我++）{    RET = pthread_create的（安培;螺纹[Ⅰ]，NULL，＆放大器; foobar_function，NULL）;    如果（保留！= 0）{
        的printf（创建pthread的错误。\\ n！）;
        出口（1）;
    }
}
 

它生成的主题线程，开始的 foobar_function 在每个。你必须（如果一切顺利的话:)）它们的ID在线程阵列。因此，例如，您可以通过调用取消第二个线程 pthread_cancel可以（线程[1]）等。

I'm trying to write a multi-threaded program, the number of threads based on command-line input, and so I can't hard-code pre-declared threads. Is this a valid way of doing it?

int threads = 5; // (dynamic, not hard-coded)
int i = 0;
pthread_t * thread = malloc(sizeof(pthread_t)*threads);

for (i = 0; i < threads; i++) {
    pthread_t foobar;
    thread[i] = foobar; // will this cause a conflict?
}

for (i = 0; i < threads; i++) {

    int ret = pthread_create(&thread[i], NULL, (void *)&foobar_function, NULL);

    if(ret != 0) {
        printf ("Create pthread error!\n");
        exit (1);
    }
}

---

Here's my result from modifications suggested below. Seems to work just fine.

int threads = 5;
int i;

pthread_t * thread = malloc(sizeof(pthread_t)*threads);

for (i = 0; i < threads; i++) {

    int ret = pthread_create(&thread[i], NULL, &foobar_function, NULL);

    if(ret != 0) {
        printf ("Create pthread error!\n");
        exit (1);
    }
    // pthread_join(thread[i], NULL); // don't actually want this here :)
}

sleep(1);     // main() will probably finish before your threads do,
free(thread); // so we'll sleep for illustrative purposes

解决方案

What's in the first cycle? Does it set the array elements to uninitialized value?

So i think that's what you need:

int threads = 5, i = 0, ret = -1;

pthread_t * thread = malloc(sizeof(pthread_t)*threads);

for (i = 0; i < threads; i++) {

    ret = pthread_create(&thread[i], NULL, &foobar_function, NULL);

    if(ret != 0) {
        printf ("Create pthread error!\n");
        exit (1);
    }
}

It spawns threads threads, starting foobar_function in each. And you have (if everything goes well:)) their ids in thread array. So for example you can cancel second thread by calling pthread_cancel(thread[1]) etc.

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