如何检索线程的“名称”?
(请参阅xcode图片,app已暂停,其中,我所谓的“名称”以黄色突出显示,“com.apple.coremedia.player.async”...我可以检索正在运行的线程,并且尝试了以下,没有运气
mach_msg_type_number_t count, i;
thread_act_array_t list;
task_threads(mach_task_self(), &list, &count);
for (i = 0; i < count; i++) {
if (list[i] == mach_thread_self()) continue;
char theName[16];
memset(theName, 0x00, sizeof(theName));
pthread_getname_np(list[i], theName);
printf("The thread name is %s.\n", theName);
}
注意:我不是要求当前线程的线程名称..我有兴趣从运行线程集中获取线程名称(参见上面的示例)..所以关于[NSThread currentThread]的解决方案不会工作
答案 0 :(得分:16)
您的问题很简单:task_threads
返回一个Mach端口数组,而不是pthread_t
数组。在致电pthread_getname_np
时,您将Mach端口视为pthread_t
。但是Mach端口不是pthread_t
。您需要使用pthread_t
pthread_from_mach_thread_np
static void dumpThreads(void) {
char name[256];
mach_msg_type_number_t count;
thread_act_array_t list;
task_threads(mach_task_self(), &list, &count);
for (int i = 0; i < count; ++i) {
pthread_t pt = pthread_from_mach_thread_np(list[i]);
if (pt) {
name[0] = '\0';
int rc = pthread_getname_np(pt, name, sizeof name);
NSLog(@"mach thread %u: getname returned %d: %s", list[i], rc, name);
} else {
NSLog(@"mach thread %u: no pthread found", list[i]);
}
}
}
我的测试程序的输出:
2013-03-14 03:21:45.908 hole[28315:c07] url connection complete
2013-03-14 03:21:46.787 hole[28315:c07] mach thread 3079: getname returned 0:
2013-03-14 03:21:46.789 hole[28315:c07] mach thread 6147: getname returned 0:
2013-03-14 03:21:46.790 hole[28315:c07] mach thread 6915: getname returned 0:
2013-03-14 03:21:46.792 hole[28315:c07] mach thread 7683: getname returned 0: WebThread
2013-03-14 03:21:46.794 hole[28315:c07] mach thread 13059: getname returned 0: com.apple.NSURLConnectionLoader
2013-03-14 03:21:46.796 hole[28315:c07] mach thread 16131: getname returned 0:
2013-03-14 03:21:46.798 hole[28315:c07] mach thread 17667: getname returned 0:
2013-03-14 03:21:46.801 hole[28315:c07] mach thread 18187: getname returned 0: com.apple.CFSocket.private
2013-03-14 03:21:46.802 hole[28315:c07] mach thread 20227: getname returned 0:
答案 1 :(得分:9)
[[NSThread currentThread] name]
应该做到这一点。