execve() 无法启动 C 程序

发布于 2024-12-20 14:51:06 字数 366 浏览 4 评论 0原文

我正在尝试在 Linux 上使用 unistd.h 中的 execve() 生成一个新进程。我尝试向其传递以下参数 execve("/bin/ls", "/bin/ls", NULL); 但没有得到结果。我也没有收到错误，程序只是退出。发生这种情况有原因吗？我尝试以 root 和普通用户身份启动它。我需要使用 execve() 的原因是因为我试图让它在程序集调用中工作，就像这样

program: db "/bin/ls",0

mov eax, 0xb
mov ebx, program
mov ecx, program
mov edx, 0
int 0x80

谢谢！

原文

I am trying to spawn a new process using execve() from unistd.h on Linux. I have tried passing it the following parameters execve("/bin/ls", "/bin/ls", NULL); but get no result. I do not get an error either, the program just exits. Is there a reason why this is happening? I have tried launching it as root and regular user. The reason I need to use execve() is because I am trying to get it to work in an assembly call like so

program: db "/bin/ls",0

mov eax, 0xb
mov ebx, program
mov ecx, program
mov edx, 0
int 0x80

Thank you!

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居里长安 2024-12-27 14:51:06

您传递给 execve 的参数是错误的。第二个和第三个必须是具有 NULL 标记值的 char 指针数组，而不是单个指针。

换句话说，就像：

#include <unistd.h>
int main (void) {
    char * const argv[] = {"/bin/ls", NULL};
    char * const envp[] = {NULL};
    int rc = execve ("/bin/ls", argv, envp);
    return rc;
}

当我运行它时，我确实得到了当前目录中的文件列表。

The arguments that you're passing to execve are wrong. Both the second and third must be an array of char pointers with a NULL sentinel value, not a single pointer.

In other words, something like:

#include <unistd.h>
int main (void) {
    char * const argv[] = {"/bin/ls", NULL};
    char * const envp[] = {NULL};
    int rc = execve ("/bin/ls", argv, envp);
    return rc;
}

When I run that, I do indeed get a list of the files in the current directory.

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梦与时光遇 2024-12-27 14:51:06

从 man 页面，

int execve(const char *filename, char *const argv[], char *const envp[]);

所以你的情况的问题是你没有通过第二个和第三个参数正确。

/* execve.c */

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

int
main(int argc, char *argv[])
{
    char *newargv[] = { NULL, "hello", "world", NULL };
    char *newenviron[] = { NULL };


newargv[0] = argv[1];

execve(argv[1], newargv, newenviron);


}
//This is a over-simplified version of the example in the man page

运行如下：

$ cc execve.c -o execve
$ ./execve ls

From the man pages,

int execve(const char *filename, char *const argv[], char *const envp[]);

So the problem in your case is that you haven't passed the 2nd and the 3rd argument correctly.

/* execve.c */

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

int
main(int argc, char *argv[])
{
    char *newargv[] = { NULL, "hello", "world", NULL };
    char *newenviron[] = { NULL };


newargv[0] = argv[1];

execve(argv[1], newargv, newenviron);


}
//This is a over-simplified version of the example in the man page

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