[英]C program to dump stack of running process on linux
I'm trying to dump the stack of a running proccess on linux fedora 20(I want to do it by myself without gdb/...) 我试图在linux fedora 20上转储正在运行的进程的堆栈(我想自己做这个没有gdb / ...)
but it looks like the default value of my buffer(buf) is never replaced even if the read() of the /proc/x/mem return me positive number. 但看起来我的缓冲区(buf)的默认值永远不会被替换,即使/ proc / x / mem的read()返回正数。
I disabled selinux 我禁用了selinux
[me@localhost ~]$ cat /proc/2419/maps
00400000-00401000 r-xp 00000000 fd:02 164622 /home/me/memo
00600000-00601000 r--p 00000000 fd:02 164622 /home/me/memo
00601000-00602000 rw-p 00001000 fd:02 164622 /home/me/memo
7fd7cbea2000-7fd7cc056000 r-xp 00000000 fd:02 8636 /usr/lib64/libc-2.18.so
7fd7cc056000-7fd7cc256000 ---p 001b4000 fd:02 8636 /usr/lib64/libc-2.18.so
7fd7cc256000-7fd7cc25a000 r--p 001b4000 fd:02 8636 /usr/lib64/libc-2.18.so
7fd7cc25a000-7fd7cc25c000 rw-p 001b8000 fd:02 8636 /usr/lib64/libc-2.18.so
7fd7cc25c000-7fd7cc261000 rw-p 00000000 00:00 0
7fd7cc261000-7fd7cc281000 r-xp 00000000 fd:02 723 /usr/lib64/ld-2.18.so
7fd7cc469000-7fd7cc46c000 rw-p 00000000 00:00 0
7fd7cc47d000-7fd7cc480000 rw-p 00000000 00:00 0
7fd7cc480000-7fd7cc481000 r--p 0001f000 fd:02 723 /usr/lib64/ld-2.18.so
7fd7cc481000-7fd7cc482000 rw-p 00020000 fd:02 723 /usr/lib64/ld-2.18.so
7fd7cc482000-7fd7cc483000 rw-p 00000000 00:00 0
7fffc8479000-7fffc849a000 rw-p 00000000 00:00 0 [stack]
7fffc854e000-7fffc8550000 r-xp 00000000 00:00 0 [vdso]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0 [vsyscall]
[me@localhost ~]$ cat analyse.c
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/ptrace.h>
#include <fcntl.h>
#define STACK_BEGIN 0x7fffc8479000
#define STACK_END 0x7fffc849a000
#define STACK_SIZE STACK_END - STACK_BEGIN
void main(int argc, char** argv)
{
int pid = atoi(argv[1]);
char mem_file_name[30];
int mem_fd;
char buf[STACK_SIZE];
sprintf(mem_file_name, "/proc/%d/mem", pid);
mem_fd = open(mem_file_name, O_RDONLY);
printf("PID=%d\n\
NAME=%s\n\
FD=%d\n\
Initial value of buf : %x\n", pid, mem_file_name, mem_fd, buf);
int ptrace_log = ptrace(PTRACE_ATTACH, pid, NULL, NULL);
waitpid(pid, NULL, 0);
int lseek_log = lseek(mem_fd, STACK_BEGIN, SEEK_SET);
int rd = read(mem_fd, buf, STACK_SIZE);
printf("in Memory at : %zd, %x (readed %d chari - ptrace ret = %d - lseek ret = %d)\n", STACK_BEGIN, buf, rd, ptrace_log, lseek_log);
ptrace(PTRACE_DETACH, pid, NULL, NULL);
close(mem_fd);
}
[me@localhost ~]$sudo ./analyse 2419
PID=2419
NAME=/proc/2419/mem
FD=3
Initial value of buf : da5030f0
in Memory at : 140736553521152, da5030f0 (readed 135168 chari - ptrace ret = 0 - lseek ret = -934834176)
read()
doesn't change the address of buf
(how could it?), it changes the contents. read()
不会改变buf
的地址(怎么可能?),它会改变内容。 If you want to see the change, you have to print the contents. 如果要查看更改,则必须打印内容。 Printing *buf
will show the first character of the buffer. Printing *buf
将显示*buf
的第一个字符。
void main(int argc, char** argv)
{
int pid = atoi(argv[1]);
char mem_file_name[30];
int mem_fd;
char buf[STACK_SIZE];
sprintf(mem_file_name, "/proc/%d/mem", pid);
mem_fd = open(mem_file_name, O_RDONLY);
printf("PID=%d\n\
NAME=%s\n\
FD=%d\n\
Initial value of buf : %c\n", pid, mem_file_name, mem_fd, *buf);
int ptrace_log = ptrace(PTRACE_ATTACH, pid, NULL, NULL);
waitpid(pid, NULL, 0);
int lseek_log = lseek(mem_fd, STACK_BEGIN, SEEK_SET);
int rd = read(mem_fd, buf, STACK_SIZE);
printf("in Memory at : %zd, %c (readed %d chari - ptrace ret = %d - lseek ret = %d)\n", STACK_BEGIN, *buf, rd, ptrace_log, lseek_log);
ptrace(PTRACE_DETACH, pid, NULL, NULL);
close(mem_fd);
}
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