如何解决/绕过似乎与GIL相关的锁定问题

Question

两个线程之间的模糊锁定似乎与Global Interpreter Lock或其他一些"幕后锁定"有关,我不知道如何继续进行故障排除.任何提示如何消除锁定将不胜感激.

该问题在更大的代码集中再现(不规律且有些随机).代码严格来说是python.Python版本是2.6.5(在Linux上).当发生锁定时,故障排除时间减少了以下问题:

1. 该程序只有两个运行线程
2. 线程同时调用受单个threading.RLock保护的方法
3. 线程1通过acquire()获取了锁[加上一些其他锁]
4. 线程2调用了acquire()并确认等待锁定
5. 线程1能够使用print()打印到控制台,但是它被一个简单的非阻塞库调用锁定

#5中的攻击性调用是函数unicode.encode,它应该是非阻塞的.线程1中线程锁定位置(如预期)中的以下代码打印'A'和'B':

print('A')
print('B')

但是,以下将只打印'A'并阻止该线程:

print('A')
u'hello'.encode('utf8') # This dummy (non-blocking) call locks up Thread 1
print('B')

这对我来说毫无意义.两个线程之间不存在逻辑死锁条件.线程1被非阻塞库调用阻塞,该调用不会以任何方式干扰线程2,线程2只是默默地等待获取RLock.我能想到线程1被阻止的唯一原因是它正在等待GIL.

有任何想法如何进一步解决这个问题,或任何机制以某种方式控制或操纵GIL操作作为一种解决方法？

编辑:响应samplebias的一些附加信息(并感谢您的回复).由于问题似乎对可能影响两个线程之间的时序的任何事情非常敏感,因此我遇到了问题.但是,运行strace只使用-f选项,经过几次迭代后,我能够得到一个跟踪.

线程1包含这三个调试语句,它们应该在控制台上打印两行"CHECK_IN"和"CHECK_TEST":

print('CHECK IN')#DEBUG
u'hello'.encode('utf8')
print('CHECK TEST')#DEBUG

这是strace的最后一页:

8605  mmap2(NULL, 266240, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb753d000
8605  mmap2(NULL, 8392704, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS|MAP_STACK, -1, 0) = 0xb6d3c000
8605  mprotect(0xb6d3c000, 4096, PROT_NONE) = 0
8605  clone(child_stack=0xb753c494, flags=CLONE_VM|CLONE_FS|CLONE_FILES|CLONE_SIGHAND|CLONE_THREAD|CLONE_SYSVSEM|CLONE_SETTLS|CLONE_PARENT_SETTID|CLONE_CHILD_CLEARTID, parent_tidptr=0xb753cbd8, {entry_number:6, base_addr:0xb753cb70, limit:1048575, seg_32bit:1, contents:0, read_exec_only:0, limit_in_pages:1, seg_not_present:0, useable:1}, child_tidptr=0xb753cbd8) = 8606
8606  set_robust_list(0xb753cbe0, 0xc <unfinished ...>
8605  futex(0xa239138, FUTEX_WAIT_PRIVATE, 0, NULL <unfinished ...>
8606  <... set_robust_list resumed> )   = 0
8606  futex(0xa239138, FUTEX_WAKE_PRIVATE, 1) = 1
8605  <... futex resumed> )             = 0
8606  gettimeofday( <unfinished ...>
8605  futex(0xa272398, FUTEX_WAIT_PRIVATE, 0, NULL <unfinished ...>
8606  <... gettimeofday resumed> {1301528807, 326496}, NULL) = 0
8606  futex(0xa272398, FUTEX_WAKE_PRIVATE, 1) = 1
8605  <... futex resumed> )             = 0
8606  futex(0xa272398, FUTEX_WAKE_PRIVATE, 1 <unfinished ...>
8605  gettimeofday( <unfinished ...>
8606  <... futex resumed> )             = 0
8605  <... gettimeofday resumed> {1301528807, 326821}, NULL) = 0
8606  futex(0xa272398, FUTEX_WAIT_PRIVATE, 0, NULL <unfinished ...>
8605  futex(0xa272398, FUTEX_WAKE_PRIVATE, 1 <unfinished ...>
8606  <... futex resumed> )             = -1 EAGAIN (Resource temporarily unavailable)
8605  <... futex resumed> )             = 0
8606  gettimeofday( <unfinished ...>
8605  futex(0xa272398, FUTEX_WAIT_PRIVATE, 0, NULL <unfinished ...>
8606  <... gettimeofday resumed> {1301528807, 326908}, NULL) = 0
8606  futex(0xa272398, FUTEX_WAKE_PRIVATE, 1) = 1
8605  <... futex resumed> )             = 0
8606  futex(0xa272398, FUTEX_WAKE_PRIVATE, 1 <unfinished ...>
8605  futex(0xa1b0d70, FUTEX_WAIT_PRIVATE, 0, NULL <unfinished ...>
8606  <... futex resumed> )             = 0
8606  stat64("/etc/localtime", {st_mode=S_IFREG|0644, st_size=2225, ...}) = 0
8606  fstat64(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 1), ...}) = 0
8606  mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xb6d3b000
8606  write(1, "CHECK IN\n", 9)         = 9
8606  futex(0xa115270, FUTEX_WAIT_PRIVATE, 0, NULL

在程序锁定之前,三行代码的输出就是以下内容:

CHECK IN

因此,strace显示了线程1(#8606)如何写入'CHECK_IN'字符串,并且当到达unicode.encode时,调用进入一个永不返回的等待状态.

顺便说一句,我将在所有模块中进行一些未来的导入以保持一些较新的python约定......

from __future__ import print_function, unicode_literals

...但是我看不出它们应该有任何区别 - 尤其是因为u'hello'字符串被明确地称为unicode字符串.

Answer 1

我在 Python 源代码中找不到任何会导致unicode.encode()阻塞的内容，并且我编写的用于尝试按预期重现此运行的虚拟程序。您提到线程 1 已获取超过 1 个锁 - 您是否已消除这些锁作为锁定源？

下面的测试用例在您的环境中是否显示出相同的锁定情况？

import time
import threading

def worker(tid):
    _lock.acquire()
    if not tid:
        # wait for rest of threads to enter acquire
        time.sleep(0.5)    
    print('%d: A' % tid)
    u'hello'.encode('utf-8')
    print('%d: B' % tid)
    _lock.release()

def start(tid):
    th = threading.Thread(target=worker, args=(tid,))
    th.start()
    return th

_num = 2
_lock = threading.RLock()
workers = [start(n) for n in range(_num)]
while all(w.isAlive() for w in workers):
    time.sleep(1)

输出：

0: A
0: B
1: A
1: B

您还可以运行strace您的程序来找出进程被阻止的位置。例如上面的脚本：

% strace -fTr -o trace.out python lockup.py

该-o trace.out标志告诉 strace 将输出写入文件。您可以省略它，strace 将打印到 stderr。

的内容trace.out应该显示程序进行的所有系统调用，每行都以线程 ID 和系统调用之间的相对时间为前缀。该行的末尾将包含该系统调用内花费的时间。我用相应的 Python 代码注释了最后几个系统调用：

# thread 0 time.sleep(0.5) completes
24778      0.500124 <... select resumed> ) = 0 (Timeout) <0.500599>
# preparing to print()
24778      0.000071 fstat(1, {st_mode=S_IFCHR|0620, st_rdev=makedev(136, 0), ...}) = 0 <0.000017>
24778      0.000058 mmap(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0x7f8fe90a6000 <0.000018>
# print("0: A\n")..
24778      0.000079 write(1, "0: A\n", 5) = 5 <0.000023>
24778      0.000106 write(1, "0: B\n", 5) = 5 <0.000056>
# thread 0 _lock.release()
24778      0.000114 futex(0xe0f3c0, FUTEX_WAKE_PRIVATE, 1) = 1 <0.000024>
24778      0.000108 madvise(0x7f8fe7266000, 8368128, MADV_DONTNEED) = 0 <0.000030>
# thread 0 exit
24778      0.000072 _exit(0)            = ?
# thread 1 _lock.acquire()
24779      0.000050 <... futex resumed> ) = 0 <0.500774>
# thread 1 print("1: A\n") and so on..
24779      0.000052 write(1, "1: A\n", 5) = 5 <0.000026>
24779      0.000086 write(1, "1: B\n", 5) = 5 <0.000026>
24779      0.000099 madvise(0x7f8fe6a65000, 8368128, MADV_DONTNEED) = 0 <0.000024>
24779      0.000064 _exit(0)            = ?
24777      0.499956 <... select resumed> ) = 0 (Timeout) <1.001138>
24777      0.000132 rt_sigaction(SIGINT, {SIG_DFL, [], SA_RESTORER, 0x7f8fe8c7c8f0}, {0x4d9a90, [], SA_RESTORER, 0x7f8fe8c7c8f0}, 8) = 0 <0.000025>
# main thread process exit
24777      0.002349 exit_group(0)       = ?