linux - 通过更改 argv[0] 更改 Linux 'ps' 输出

Question

我正在尝试让一个程序更改“ps”显示为进程的 CMD 名称的内容，使用我所推荐的技术来简单地覆盖 argv[0] 指向的内存。这是我写的示例程序。

#include <iostream>
#include <cstring>
#include <sys/prctl.h>
#include <linux/prctl.h>
using std::cout;
using std::endl;
using std::memcpy;

int main(int argc, char** argv)  {
    if ( argc < 2 ) {
        cout << "You forgot to give new name." << endl;
        return 1;
    }

    // Set new 'ps' command name - NOTE that it can't be longer than
    // what was originally in argv[0]!

    const char *ps_name = argv[1];
    size_t arg0_strlen = strlen(argv[0]);
    size_t ps_strlen = strlen(ps_name);
    cout << "Original argv[0] is '" << argv[0] << "'" << endl;

    // truncate if needed
    size_t copy_len = (ps_strlen < arg0_strlen) ? ps_strlen+1 : arg0_strlen;
    memcpy((void *)argv[0], ps_name, copy_len);
    cout << "New name for ps is '" << argv[0] << "'" << endl;

    cout << "Now spin.  Go run ps -ef and see what command is." << endl;
    while (1) {};
}

输出是:

$ ./ps_test2 foo
Original argv[0] is './ps_test2'
New name for ps is 'foo'
Now spin.  Go run ps -ef and see what command is.

ps -ef 的输出是:

5079     28952  9142 95 15:55 pts/20   00:00:08 foo _test2 foo

显然，插入了“foo”，但它的空终止符要么被忽略，要么变成空白。原始 argv[0] 的尾部仍然可见。

如何替换“ps”打印的字符串？

Answer 1

您需要重写整个命令行，它在 Linux 中存储为一个连续的缓冲区，参数以零分隔。

类似于:

size_t cmdline_len = argv[argc-1] + strlen(argv[argc-1]) - argv[0];
size_t copy_len = (ps_strlen + 1 < cmdline_len) ? ps_strlen + 1 : cmdline_len;
memcpy(argv[0], ps_name, copy_len);
memset(argv[0] + copy_len, 0, cmdline_len - copy_len);