clone()一调用，程序状态就异常了
本帖最后由 cdsfiui 于 2017-01-11 15:38 编辑如下代码，我想看clone()是不是能创建新进程，而且让两个进程都能正常运行和打印#include<stdio.h>
#include<sched.h>
#include<unistd.h>
#include<sys/types.h>
#include<errno.h>
int f(void*arg)
{
pid_t pid=getpid();
printf("child pid=%d\n",pid);
}
char buf;
int main()
{
printf("before clone\n");
int pid=clone(f,buf,CLONE_VM|CLONE_VFORK,NULL);
if(pid==-1){
printf("%d\n",errno);
return 1;
}
waitpid(pid,NULL,0);
printf("after clone\n");
printf("father pid=%d\n",getpid());
return 0;
}
编译运行的结果是:
$g++ testClone.cpp && ./a.out
before clone
很奇怪，clone之后，程序就进入了异常状态
我尝试gdb，调试到clone只一句执行以后，有如下输出：
Breakpoint 1, main () at testClone.cpp:15
(gdb) n-
before clone
(gdb) n-
waiting for new child: No child processes.
(gdb) n-
Single stepping until exit from function clone@plt,-
which has no line number information.
如果我去掉waitpid这一句，那么运行结果一样，gdb的输出变成了:
(gdb) n-
before clone
(gdb) n-
Detaching after fork from child process 26709.
warning: Unexpected waitpid result 000000 when waiting for vfork-done
Cannot remove breakpoints because program is no longer writable.
It might be running in another process.
Further execution is probably impossible.
0x00007fb18a446bf1 in clone () from /lib64/libc.so.6
ptrace: No such process.
我的程序错在哪里了? 谢谢
两个进程要加锁 我知道了，clone第二个参数必须是指向栈底的指针，数组首地址是栈顶，反了，所以改成:
char buf; // *** allocate more stack ***
int main()
{
printf("before clone\n");
int pid=clone(f,buf+sizeof(buf),CLONE_VM|CLONE_VFORK,NULL);
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