在连接线程上调用pthread_cancel会导致linux下的segfault

Question

The following code ends with a segmentation fault on the first call to pthread_cancel but only under linux. Under Mac OS it runs fine. Am I not allowed to call pthread_cancel on a thread that has finished running? Maybe I should not call pthread_cancel at all?

#include <iostream>
#include <pthread.h>

using namespace std;


void* run(void *args) {
   cerr << "Hallo, Running" << endl;
}   

int main() {
    int n = 100;
    pthread_t* pool = new pthread_t[n];

    for(int i=0;i<n;i++) {
        pthread_t tmp;
        pthread_create(&tmp,NULL,&run,NULL);
        pool[i] = (tmp);
    }

    for(int i=0;i<n;i++) {
        pthread_join(pool[i],0);
    }

    for(int i=0;i<n;i++) {
        pthread_cancel(pool[i]);
    }
}

Answer 1

See POSIX XSH 2.9.2 :

Although implementations may have thread IDs that are unique in a system, applications should only assume that thread IDs are usable and unique within a single process. The effect of calling any of the functions defined in this volume of POSIX.1-2008 and passing as an argument the thread ID of a thread from another process is unspecified. The lifetime of a thread ID ends after the thread terminates if it was created with the detachstate attribute set to PTHREAD_CREATE_DETACHED or if pthread_detach() or pthread_join() has been called for that thread. A conforming implementation is free to reuse a thread ID after its lifetime has ended. If an application attempts to use a thread ID whose lifetime has ended, the behavior is undefined.

If a thread is detached, its thread ID is invalid for use as an argument in a call to pthread_detach() or pthread_join().

You may not use a pthread_t after the thread it refers to has been joined, or if the thread has terminated while detached. Simply remove the pthread_cancel code from your program. It's wrong. pthread_cancel is for cancelling an in-progress thread, and has very tricky requirements for using it safely without causing resource leaks. It's not useful for threads which exit on their own.