用glibc打电话给gettid

问题描述

我正在glibc中工作，我需要获取当前线程的ID.为此，我用syscall(SYS_gettid);问题是，我被迫加入bits/syscall.h而不是理想情况，即sys/syscall.h.

I am working in glibc and I need to get the id of the current thread. For this i use syscall(SYS_gettid); Issue is, i am forced to include bits/syscall.h instead of ideal case i.e sys/syscall.h.

sys/syscall.h内部调用bits/syscall.h，但是用#ifndef _LIBC宏包装.即

sys/syscall.h internally calls bits/syscall.h but that is wrapped with #ifndef _LIBC macro. i.e

     #ifndef _LIBC
        /* The Linux kernel header file defines macros `__NR_<name>', but some
           programs expect the traditional form `SYS_<name>'.  So in building libc
           we scan the kernel's list and produce <bits/syscall.h> with macros for
           all the `SYS_' names.  */
       # include <bits/syscall.h>
    #endif

也bits/syscall.h指出 切勿直接使用bits/syscall.h；而应使用sys/syscall.h."

also bits/syscall.h states that "Never use bits/syscall.h directly; include sys/syscall.h instead."

由于在我的情况下将定义_LIBC，因为我直接在malloc.c中编写代码，请提出我该如何克服这个问题.

Since _LIBC will be defined in my case as i am writing code directly in malloc.c, please suggest my how can i overcome this.

谢谢，卡皮尔

推荐答案

gettid()是系统调用.就我所知，没有gettid的glibc包装器.您需要使用syscall()调用gettid().以下代码对我有用.

gettid() is a system call. As for as I know there is no glibc wrapper for gettid. You need to invoke gettid() using syscall(). The following code works for me.

#include <sys/syscall.h>
#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>

int main()
{
    long tid;

    tid = syscall(SYS_gettid);
    printf("%ld\n", tid);
    return EXIT_SUCCESS;
}

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