子线程退出时如何通知父线程

有人可以帮助我理解子线程退出后唤醒主线程的情况吗？我的理解是，联接调用futex_wait，导致主线程等待，直到子线程完成。大概有一个调用futex_wake允许主线程恢复，但在两个线程上运行strace后，我无法看到系统调用唤醒父线程的位置。子线程退出时如何通知父线程

#include <iostream> 
#include <thread> 
#include <string> 

void hello() 
{ 
    std::cout <<"Hello Concurrent World" << std::endl; 
} 

int main() 
{ 
    std::thread t(hello); 
    t.join(); 

    std::cout << "finished" << std::endl; 
}

从父线程strace的

19:37:06.206727 brk(NULL)    = 0x1380000 
19:37:06.206804 brk(0x13b2000)   = 0x13b2000 
19:37:06.206968 futex(0x7fcf7953005c, FUTEX_WAKE_PRIVATE, 2147483647) = 0 
19:37:06.207331 futex(0x7fcf79530068, FUTEX_WAKE_PRIVATE, 2147483647) = 0 
19:37:06.207577 mmap(NULL, 8392704, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS|MAP_STACK, -1, 0) = 0x7fcf780ba000 
19:37:06.207726 mprotect(0x7fcf780ba000, 4096, PROT_NONE) = 0 
19:37:06.207863 clone(child_stack=0x7fcf788b9fb0, flags=CLONE_VM|CLONE_FS|CLONE_FILES|CLONE_SIGHAND|CLONE_THREAD|CLONE_SYSVSEM|CLONE_SETTLS|CLONE_PARENT_SETTID|CLONE_CHILD_CLEARTID, parent_tidptr=0x7fcf788ba9d0, tls=0x7fcf788ba700, child_tidptr=0x7fcf788ba9d0) = 10204 
19:37:06.208005 futex(0x7fcf788ba9d0, FUTEX_WAIT, 10204, NULL) = 0 
19:37:06.209781 write(1, "finished\n", 9) = 9 
19:37:06.210021 exit_group(0)   = ? 
19:37:06.210394 +++ exited with 0 +++

从子线程strace的

19:07:09.837174 set_robust_list(0x7f123adba9e0, 24) = 0 
19:07:09.837411 fstat(1, {st_mode=S_IFIFO|0600, st_size=0, ...}) = 0 
19:07:09.837508 mmap(NULL, 134217728, PROT_NONE, MAP_PRIVATE|MAP_ANONYMOUS|MAP_NORESERVE, -1, 0) = 0x7f12325ba000 
19:07:09.837601 munmap(0x7f12325ba000, 27549696) = 0 
19:07:09.837693 munmap(0x7f1238000000, 39559168) = 0 
19:07:09.837818 mprotect(0x7f1234000000, 139264, PROT_READ|PROT_WRITE) = 0 
19:07:09.837947 write(1, "Hello Concurrent World\n", 23) = 23 
19:07:09.838082 madvise(0x7f123a5ba000, 8368128, MADV_DONTNEED) = 0 
19:07:09.838164 exit(0)     = ? 
19:07:09.838285 +++ exited with 0 +++

目前使用Ubuntu 16.10，铛版本3.8.1-12，内核编译4.8.0-34-通用

来源

2017-02-05 js7222

此：'19：37：06.208005 futex的（0x7fcf788ba9d0，FUTEX_WAIT，10204，NULL）= 0'看上去非常像一个等待。 –

我看到父线程正在等待。我不明白的是哪个系统调用唤醒父线程来通知子线程已经完成。 – js7222

此调用：set_robust_list(0x7f123adba9e0, 24) = 0设置了一个健壮的futex，它会在线程退出时自动恢复。所以不需要系统调用来唤醒futex。内核将在线程退出时唤醒futex（即使线程被终止）。

下面是从Debian的拉伸迹片段表明该fuxes恢复 “本身”：

[pid 21604] 20:38:49.614788 futex(0x7f0349dbc9d0, FUTEX_WAIT, 21605, NULL <unfinished ...> 
[pid 21605] 20:38:49.614851 set_robust_list(0x7f0349dbc9e0, 24) = 0 
[pid 21605] 20:38:49.614948 nanosleep({tv_sec=0, tv_nsec=100000000}, NULL) = 0 
[pid 21605] 20:38:49.715377 madvise(0x7f03495bc000, 8368128, MADV_DONTNEED) = 0 
[pid 21605] 20:38:49.715554 exit(0)  = ? 
[pid 21604] 20:38:49.715674 <... futex resumed>) = 0 
[pid 21605] 20:38:49.715701 +++ exited with 0 +++

来源

2017-02-05 20:44:55 rustyx

子线程退出时如何通知父线程

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