如何在Linux上使用clone()创建真正的线程?
我正在尝试使用
clone()创建一个新线程.使用以下代码(…):
#include <stdio.h> #include <stdlib.h> #include <unistd.h> #define _SCHED_H 1 #define __USE_GNU 1 #include <bits/sched.h> #define STACK_SIZE 4096 int func(void *arg) { printf("Inside func.n"); sleep(1); printf("Terminating func...n"); return 0; } int main() { printf("This process pid: %un",getpid()); char status_file[] = "/proc/self/status"; void *child_stack = malloc(STACK_SIZE); int thread_pid; printf("Creating new thread...n"); thread_pid = clone(&func,child_stack+STACK_SIZE,CLONE_SIGHAND|CLONE_FS|CLONE_VM|CLONE_FILES,NULL); printf("Done! Thread pid: %dn",thread_pid); FILE *fp = fopen(status_file,"rb"); printf("Looking into %s...n",status_file); while(1) { char ch = fgetc(fp); if(feof(fp)) break; printf("%c",ch); } fclose(fp); getchar(); return 0; } 我得到以下内容: This process pid: 10839 Creating new thread... Done! Thread pid: 10840 Inside func. Looking into /proc/self/status... Name: threadTest02 State: R (running) Tgid: 10839 Pid: 10839 PPid: 4777 TracerPid: 0 Uid: 1000 1000 1000 1000 Gid: 1000 1000 1000 1000 FDSize: 256 Groups: 4 20 24 27 30 46 107 123 124 1000 VmPeak: 4300 kB VmSize: 4300 kB VmLck: 0 kB VmPin: 0 kB VmHWM: 356 kB VmRSS: 356 kB VmData: 188 kB VmStk: 136 kB VmExe: 4 kB VmLib: 1884 kB VmPTE: 32 kB VmSwap: 0 kB Threads: 1 SigQ: 0/22869 SigPnd: 0000000000000000 ShdPnd: 0000000000000000 SigBlk: 0000000000000000 SigIgn: 0000000000000000 SigCgt: 0000000000000000 CapInh: 0000000000000000 CapPrm: 0000000000000000 CapEff: 0000000000000000 CapBnd: ffffffffffffffff Cpus_allowed: 3 Cpus_allowed_list: 0-1 Mems_allowed: 00000000,00000001 Mems_allowed_list: 0 voluntary_ctxt_switches: 1 nonvoluntary_ctxt_switches: 1 Terminating func... 那么,简而言之,我的计划做了什么?它使用clone创建一个新线程并打印其/ proc / self / status,以便我可以看到它的状态.由于我的线程休眠1秒,因此在打印/ proc / self / status时它仍处于活动状态. 但是,至少有两件事使我的线程不像普通线程.首先,正如您在上面看到的,进程’pid是10839而我的线程的pid是10840.因此,进程和我的线程没有相同的pid,就像在一个共同的线程中一样.其次,即使在我的线程创建之后,我的进程’/ proc / self / status文件的Threads:字段仍然是1.所以,我的线程似乎不被识别为线程. 我的问题是,我的代码中缺少什么?我需要做些什么才能使我的线程像普通线程一样?克隆的第三个参数中是否缺少任何选项? 解决方法
您可能希望看到标志CLONE_THREAD,它会将新线程放在与调用进程相同的线程组中.
一旦你给了CLONE_THREAD,它将使新线程与调用进程具有相同的pid和ppid.它用于posix线程.以下是我系统的输出. LWP专栏说这是一个轻量级的过程,具有不同的TID UID PID PPID LWP C NLWP SZ RSS PSR STIME TTY TIME CMD anukalp 18398 9638 18398 0 2 464 456 0 10:56 pts/3 00:00:00 ./a.out anukalp 18398 9638 18399 0 2 464 456 1 10:56 pts/3 00:00:00 ./a.out 此外,/ proc / self / status的输出发生了变化,我添加了几个printfs: [anukalp @ localhost~] $./a.out This process pid: 18398 Creating new thread... Done! Thread pid: 18399 /* This is now thread id,available to caller of clone */ getpid(): ad pid: 18399 Inside func. getpid(): 18398 getppid(): 9638 Looking into /proc/self/status... Name: a.out State: R (running) Tgid: 18398 Pid: 18398 PPid: 9638 TracerPid: 0 Uid: 500 500 500 500 Gid: 500 500 500 500 FDSize: 256 Groups: 7 19 22 80 81 82 83 100 490 500 VmPeak: 1856 kB VmSize: 1856 kB VmLck: 0 kB VmPin: 0 kB VmHWM: 248 kB VmRSS: 248 kB VmData: 168 kB VmStk: 140 kB VmExe: 4 kB VmLib: 1516 kB VmPTE: 16 kB VmSwap: 0 kB Threads: 2 SigQ: 1/14050 SigPnd: 0000000000000000 ShdPnd: 0000000000000000 SigBlk: 0000000000000000 SigIgn: 0000000000000000 SigCgt: 0000000000000000 CapInh: 0000000000000000 CapPrm: 0000000000000000 CapEff: 0000000000000000 CapBnd: ffffffffffffffff Cpus_allowed: 00000000,000000ff Cpus_allowed_list: 0-7 voluntary_ctxt_switches: 1 nonvoluntary_ctxt_switches: 0 Inside thread: thread pid = 18398 Inside thread: thread ppid = 9638 请让我知道这可不可以帮你! (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |