本文起初主要想写个演示样例实測下pthread_mutex_lock和pthread_mutex_trylock差别。在linux机器上非常快就over了，可是想了一下。pthread是unix系的，在windows没办法直接执行代码非常不方便。于是想到了android，windows上安装ndk，手机root就能够跑pthread代码咯。。。
demo
lock和trylock的差别也非常好理解：，前者是堵塞的。死等知道相互排斥锁被释放；而后者则更加灵活。浅尝辄止。做个尝试不行则干其它事情去。。測试代码pt_lock.c例如以下：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
#include <unistd.h>
#include <sys/types.h>
typedef pthread_t pt_t;
typedef unsigned int uint_t;
pthread_mutex_t mt = PTHREAD_MUTEX_INITIALIZER;
void lock_func(void* arg)
{
pid_t pid; //process
pt_t tid; // thread
pid = getpid();
tid = pthread_self();
printf("want to lock mutex, msg=%s, tid=%u
", (char*)arg, (uint_t)tid);
pthread_mutex_lock( &mt );
printf("I[tid=%u] am using, (*|^_^|*)
", (uint_t)tid);
sleep(10);
pthread_mutex_unlock( &mt );
}
void try_lock_func(void* arg)
{
uint_t tid = (uint_t)pthread_self();
int counter = 0;
while ( pthread_mutex_trylock( &mt ) )
{
sleep(1);
++counter;
printf("after sleep 1s, i [tid=%u] want to try again, iter=%d.
", tid, counter);
}
printf("It is my[tid=%u] turn, so long i waited...msg=%s
", tid, (char*)arg);
pthread_mutex_unlock( &mt );
}
#define XX_CREATE_FAILED(err)
printf("create thread error : %s
", strerror(err));
return 1;
int main()
{
int rc;
pt_t pt1, pt2, pt3;
const char* msg1 = "block";
const char* msg2 = "unblock";
rc = pthread_create(&pt1, NULL, (void*)&lock_func, (void*)msg1); if (rc != 0)
{
XX_CREATE_FAILED(rc);
}
rc = pthread_create(&pt2, NULL, (void*)&lock_func, (void*)msg1); if (rc != 0)
{
XX_CREATE_FAILED(rc);
}
sleep(1);
rc = pthread_create(&pt3, NULL, (void*)&try_lock_func, (void*)msg2); if (rc != 0)
{
XX_CREATE_FAILED(rc);
}
pthread_join(pt1, NULL);
pthread_join(pt2, NULL);
pthread_join(pt3, NULL);
return 0;
}
代码思路也非常好理解：创建三个线程。1和2通过lock方式去争抢mt相互排斥锁，3线程则灵活，每隔1秒去检測下mt相互排斥锁能否够用，不会堵塞。线程1或者2占有mt的时间为10秒。
linux run
xx@h10-1-152-72:~/peteryfren/cpp/pthread> gcc -Wall -o pt_lock pt_lock.c -lpthread xx@h10-1-152-72:~/peteryfren/cpp/pthread> ./pt_lock输出结果与预期一致，id=1082132800线程先占有mt。10s内线程tid=1090525504堵塞，而线程tid=1098918208每隔1s測试下mt可用性。
want to lock mutex, msg=block, tid=1082132800 I[tid=1082132800] am using, (*|^_^|*) want to lock mutex, msg=block, tid=1090525504 after sleep 1s, i [tid=1098918208] want to try again, iter=1. after sleep 1s, i [tid=1098918208] want to try again, iter=2. after sleep 1s, i [tid=1098918208] want to try again, iter=3. after sleep 1s, i [tid=1098918208] want to try again, iter=4. after sleep 1s, i [tid=1098918208] want to try again, iter=5. after sleep 1s, i [tid=1098918208] want to try again, iter=6. after sleep 1s, i [tid=1098918208] want to try again, iter=7. after sleep 1s, i [tid=1098918208] want to try again, iter=8. I[tid=1090525504] am using, (*|^_^|*) after sleep 1s, i [tid=1098918208] want to try again, iter=9. after sleep 1s, i [tid=1098918208] want to try again, iter=10. after sleep 1s, i [tid=1098918208] want to try again, iter=11. after sleep 1s, i [tid=1098918208] want to try again, iter=12. after sleep 1s, i [tid=1098918208] want to try again, iter=13. after sleep 1s, i [tid=1098918208] want to try again, iter=14. after sleep 1s, i [tid=1098918208] want to try again, iter=15. after sleep 1s, i [tid=1098918208] want to try again, iter=16. after sleep 1s, i [tid=1098918208] want to try again, iter=17. after sleep 1s, i [tid=1098918208] want to try again, iter=18. after sleep 1s, i [tid=1098918208] want to try again, iter=19. It is my[tid=1098918208] turn, so long i waited...msg=unblock
android run
android上编译和执行pt_lock參考前一篇blog：http://blog.csdn.net/ryfdizuo/article/details/28891649 详细批处理run.cmd例如以下：
@echo "1. build .o file" @Rem fPIE flag is used in compiling stage. D:android-ndk-r9b-windows-x86 oolchainsarm-linux-androideabi-4.6prebuiltwindowsinarm-linux-androideabi-gcc.exe --sysroot=D:android-ndk-r9b-windows-x86platformsandroid-13arch-arm -fPIE -c pt_lock.c @echo "2. build exec file" @Rem pie flag is used in linking stage. D:android-ndk-r9b-windows-x86 oolchainsarm-linux-androideabi-4.6prebuiltwindowsinarm-linux-androideabi-gcc.exe --sysroot=D:android-ndk-r9b-windows-x86platformsandroid-13arch-arm -pie -o pt_lock pt_lock.o @pause里面的gcc和android系统库路径须要依据机器上实际位置改动。
注意android上gcc编译与linux上有点点区别：ndk中gcc不须要须要显式指定-lpthread。可能pthread默认就会链接。
android-ndk-r9b-windows-x86platformsandroid-13arch-armusrlib文件夹下的这些so载入须要指定。。在Android.mk中类似。
命令行下执行run.cmd，编译pt_lock，然后推到手机上，注意我的nexus4必须是/data/local/tmp以下才有权限，执行后有错误，详细步骤：
E:GitHub dk_tutorialpthread_playground>adb shell root@mako:/ # cd /data/local/tmp cd /data/local/tmp root@mako:/data/local/tmp # mv /sdcard/pt_lock ./ mv /sdcard/pt_lock ./ root@mako:/data/local/tmp # chmod 751 pt_lock chmod 751 pt_lock root@mako:/data/local/tmp # ./pt_lock error: only position independent executables (PIE) are supported.报了PIE不支持的错误，n4上安装的是android L系统。gg后知道编译的链接的须要添加pie等设置，參见[ref2]，fPIE是编译时候的选项，pie是链接时候的选项。再次又一次编译，推到手机上执行OK。。。pt_lock.c不须要做不论什么改动。
pthread学习
1.线程相关
1）回调函数类型 void* (*pfunc) (void* arg) 。參数通过void*传入，多个參数是通过结构体/数组打包。
2）线程属性，能够设置线程类型，通常是joinable。设置线程栈的大小。假设线程内部要定义大数组一定小心栈越界。
3）线程终止的几种情况：
- 回调函数执行完毕。正常退出
- 线程内调用pthread_exit函数。无论是否完毕都退出。该函数后面的全部代码都不会被执行（尤其printf等语句）。
- 其它线程显式调用pthread_cancel 结束当前线程。
- 整个进程被终止，由于调用了exit等函数。它以下的全部线程都被杀死。资源被回收。
- main函数创建N个线程。两种做法等待其它线程结束：显式pthread_join等待其它线程结束；main函数中调用pthread_exit，等待其它线程结束。
2. 相互排斥锁相关
pthread_mutex_init
pthread_mutex_lock。堵塞等待
pthread_mutex_trylock，非堵塞等待
pthread_mutex_destroy
3. 信号量相关
pthread_mutex_init
pthread_mutex_wait 堵塞当前线程，直到满足某些条件时。伪代码例如以下：
begin
pthread_mutex_unlock( mutex ) 堵塞之前释放相互排斥锁
block_on_cond( condition ) 条件堵塞
pthread_mutex_lock(mutex) 条件满足后 占有相互排斥锁，完毕工作后须要释放相互排斥锁
end
该语句必须在循环中运行，而不是if中，由于线程之间可能有虚假的唤醒行为，spurious wakeup。当唤醒时须要再次确定预期的条件是否满足。假设不满足继续等待。
相关的讨论參见：http://a-shi.org/blog/2012/07/27/pthread_cond_wait_mutex_while/
pthread_mutex_signal，仅仅有一个线程等待时发送信号量
pthread_mutex_broadcast 多个线程同一时候等待时使用
pthread_mutex_destroy
4. 生存者/消费者，读写。哲学家进餐模型
refer
1. gcc以下的pie和fPIE 选项标志解说，http://richardustc.github.io/blog/2013/05/pie/
2. android L系统下PIE错误，http://blog.csdn.net/hxdanya/article/details/39371759
3. pthread http://man7.org/linux/man-pages/man3/pthread_create.3.html