清单 2. 编译测试程序
[dyu@xilinuxbldsrvpurify]$ g++ -g lock.cpp -o lock -lpthread
清单 3. 查找测试程序的进程号
[dyu@xilinuxbldsrvpurify]$ ps -ef|greplock
dyu 6721 5751 0 15:21 pts/3 00:00:00 ./lock
清单 4. 对死锁进程第一次执行 pstack（pstack –进程号）的输出结果
[dyu@xilinuxbldsrvpurify]$ pstack 6721 Thread 5 (Thread 0x41e37940 (LWP 6722)): #0 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 #1 0x0000003d1a808e1a in _L_lock_1034 () from /lib64/libpthread.so.0 #2 0x0000003d1a808cdc in pthread_mutex_lock () from /lib64/libpthread.so.0 #3 0x0000000000400a9b in func1() () #4 0x0000000000400ad7 in thread1(void*) () #5 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #6 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 4 (Thread 0x42838940 (LWP 6723)): #0 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 #1 0x0000003d1a808e1a in _L_lock_1034 () from /lib64/libpthread.so.0 #2 0x0000003d1a808cdc in pthread_mutex_lock () from /lib64/libpthread.so.0 #3 0x0000000000400a17 in func2() () #4 0x0000000000400a53 in thread2(void*) () #5 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #6 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 3 (Thread 0x43239940 (LWP 6724)): #0 0x0000003d19c9a541 in nanosleep () from /lib64/libc.so.6 #1 0x0000003d19c9a364 in sleep () from /lib64/libc.so.6 #2 0x00000000004009bc in thread3(void*) () #3 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #4 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 2 (Thread 0x43c3a940 (LWP 6725)): #0 0x0000003d19c9a541 in nanosleep () from /lib64/libc.so.6 #1 0x0000003d19c9a364 in sleep () from /lib64/libc.so.6 #2 0x0000000000400976 in thread4(void*) () #3 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #4 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 1 (Thread 0x2b984ecabd90 (LWP 6721)): #0 0x0000003d1a807b35 in pthread_join () from /lib64/libpthread.so.0 #1 0x0000000000400900 in main () |
清单 5. 对死锁进程第二次执行 pstack（pstack –进程号）的输出结果
[dyu@xilinuxbldsrvpurify]$ pstack 6721 Thread 5 (Thread 0x40bd6940 (LWP 6722)): #0 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 #1 0x0000003d1a808e1a in _L_lock_1034 () from /lib64/libpthread.so.0 #2 0x0000003d1a808cdc in pthread_mutex_lock () from /lib64/libpthread.so.0 #3 0x0000000000400a87 in func1() () #4 0x0000000000400ac3 in thread1(void*) () #5 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #6 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 4 (Thread 0x415d7940 (LWP 6723)): #0 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 #1 0x0000003d1a808e1a in _L_lock_1034 () from /lib64/libpthread.so.0 #2 0x0000003d1a808cdc in pthread_mutex_lock () from /lib64/libpthread.so.0 #3 0x0000000000400a03 in func2() () #4 0x0000000000400a3f in thread2(void*) () #5 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #6 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 3 (Thread 0x41fd8940 (LWP 6724)): #0 0x0000003d19c7aec2 in memset () from /lib64/libc.so.6 #1 0x00000000004009be in thread3(void*) () #2 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #3 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 2 (Thread 0x429d9940 (LWP 6725)): #0 0x0000003d19c7ae0d in memset () from /lib64/libc.so.6 #1 0x0000000000400982 in thread4(void*) () #2 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #3 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 Thread 1 (Thread 0x2af906fd9d90 (LWP 6721)): #0 0x0000003d1a807b35 in pthread_join () from /lib64/libpthread.so.0 #1 0x0000000000400900 in main () |
连续多次查看这个进程的函数调用关系堆栈进行分析：当进程吊死时，多次使用 pstack 查看进程的函数调用堆栈，死锁线程将一直处于等锁的状态，对比多次的函数调用堆栈输出结果，确定哪两个线程（或者几个线程）一直没有变化且一直处于等锁的状态（可能存在两个线程 一直没有变化）。
输出分析：
根据上面的输出对比可以发现，线程 1 和线程 2 由第一次 pstack 输出的处在 sleep 函数变化为第二次 pstack 输出的处在 memset 函数。但是线程 4 和线程 5 一直处在等锁状态（pthread_mutex_lock），在连续两次的 pstack 信息输出中没有变化，所以我们可以推测线程 4 和线程 5 发生了死锁。
Gdb into thread 输出：
清单 6. 然后通过 gdb attach 到死锁进程
(gdb) infothread 5 Thread 0x41e37940 (LWP 6722) 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 4 Thread 0x42838940 (LWP 6723) 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 3 Thread 0x43239940 (LWP 6724) 0x0000003d19c9a541 in nanosleep () from /lib64/libc.so.6 2 Thread 0x43c3a940 (LWP 6725) 0x0000003d19c9a541 in nanosleep () from /lib64/libc.so.6 * 1 Thread 0x2b984ecabd90 (LWP 6721) 0x0000003d1a807b35 in pthread_join () from /lib64/libpthread.so.0 |
清单 7. 切换到线程 5 的输出
(gdb) thread 5 [Switchingto thread 5 (Thread 0x41e37940 (LWP 6722))]#0 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 (gdb) where #0 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 #1 0x0000003d1a808e1a in _L_lock_1034 () from /lib64/libpthread.so.0 #2 0x0000003d1a808cdc in pthread_mutex_lock () from /lib64/libpthread.so.0 #3 0x0000000000400a9b in func1 () at lock.cpp:18 #4 0x0000000000400ad7 in thread1 (arg=0x0) at lock.cpp:43 #5 0x0000003d1a80673d in start_thread () from /lib64/libpthread.so.0 #6 0x0000003d19cd40cd in clone () from /lib64/libc.so.6 |
清单 8. 线程 4 和线程 5 的输出
(gdb) f 3 #3 0x0000000000400a9b in func1 () at lock.cpp:18 18 pthread_mutex_lock(&mutex2); (gdb) thread 4 [Switchingto thread 4 (Thread 0x42838940 (LWP 6723))]#0 0x0000003d1a80d4c4 in __lll_lock_wait () from /lib64/libpthread.so.0 (gdb) f 3 #3 0x0000000000400a17 in func2 () at lock.cpp:31 31 pthread_mutex_lock(&mutex1); (gdb) p mutex1 $1 = {__data = {__lock = 2, __count = 0, __owner = 6722, __nusers = 1, __kind = 0, __spins = 0, __list = {__prev = 0x0, __next = 0x0}}, __size = "0200000000000000B32000001", '00' , __align = 2} (gdb) p mutex3 $2 = {__data = {__lock = 0, __count = 0, __owner = 0, __nusers = 0, __kind = 0, __spins = 0, __list = {__prev = 0x0, __next = 0x0}}, __size = '00' , __align = 0} (gdb) p mutex2 $3 = {__data = {__lock = 2, __count = 0, __owner = 6723, __nusers = 1, __kind = 0, __spins = 0, __list = {__prev = 0x0, __next = 0x0}}, __size = "0200000000000000C32000001", '00' , __align = 2} (gdb) |
从上面可以发现，线程 4 正试图获得锁 mutex1，但是锁 mutex1 已经被 LWP 为 6722 的线程得到（__owner = 6722），线程 5 正试图获得锁 mutex2，但是锁 mutex2 已经被 LWP 为 6723 的 得到（__owner = 6723），从 pstack 的输出可以发现，LWP 6722 与线程 5 是对应的，LWP 6723 与线程 4 是对应的。所以我们可以得出， 线程 4 和线程 5 发生了交叉持锁的死锁现象。查看线程的源代码发现，线程 4 和线程 5 同时使用 mutex1 和 mutex2，且申请顺序不合理。
总结
本文简单介绍了一种在 Linux 平台下分析死锁问题的方法，对一些死锁问题的分析有一定作用。希望对大家有帮助。理解了死锁的原因，尤其是产生死锁的四个必要条件，就可以最大可能地避免、预防和解除死锁。所以，在系统设计、进程调度等方面注意如何不让这四个必要条件成立，如何确定资源的合理分配算法，避免进程永久占据系统资源。此外，也要防止进程在处于等待状态的情况下占用资源 , 在系统运行过程中，对进程发出的每一个系统能够满足的资源申请进行动态检查，并根据检查结果决定是否分配资源，若分配后系统可能发生死锁，则不予分配，否则予以分配。因此，对资源的分配要给予合理的规划，使用有序资源分配法和银行家算法等是避免死锁的有效方法。
