使用动态参数列表调用 ctypes.execve

Question

我想在python中使用ctypes.execve来调用外部脚本。我已经弄清楚如何调用脚本以及如何传递参数列表。 但正如你所看到的，列表或参数是相当静态的，我不知道如何动态地做到这一点。

参数列表应为Python列表 例如：

argv_list = ["./tmp/arg_test.sh", "arg_1", "arg_2"]

这个Python列表需要以某种方式转换以匹配

char * const argv[]

，这样我就可以将它传递给execve。 <-- This is where I am struggling.

cmd = "./tmp/arg_test.sh"
cmd_byte = cmd.encode('utf-8')     # create byte objects from the strings

### works ### ### works ### ### works ### ### works ###

argv_0 = "argv_0"
argv_0_byte = argv_0.encode('utf-8')

argv_1 = "argv_1"
argv_1_byte = argv_1.encode('utf-8')

argv_2 = "argv_2"
argv_2_byte = argv_2.encode('utf-8')

argv_3 = "argv_3"
argv_3_byte = argv_3.encode('utf-8')

b_argv = (ctypes.c_char_p * 5)(argv_0_byte, argv_1_byte, argv_2_byte, argv_3_byte)

### works ### ### works ### ### works ### ### works ###


# int execve(const char * filename, char * const argv[], char * const envp[]);
libc.execve(ctypes.c_char_p(cmd_byte), b_argv, 0)

谢谢！

Answer 1

这是一种动态生成参数的方法。这是为 Windows

msvcrt

编写的，但也应该适用于

libc

：

测试.cpp

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

int main(int argc, char* argv[])
{
    for(int i = 0; i < argc; ++i)
        printf("argv[%d] = \"%s\"\n", i, argv[i]);
    printf("TEST1=%s\n", getenv("TEST1"));
    printf("TEST2=%s\n", getenv("TEST2"));
    return 0;
}

测试.py

import ctypes as ct

dll = ct.CDLL('msvcrt')
dll._execve.argtypes = ct.c_char_p, ct.POINTER(ct.c_char_p), ct.POINTER(ct.c_char_p)
dll._execve.restype = ct.c_ssize_t

cmd = b'test.exe'
args = [b'arg1', b'arg2', b'arg3']
env = [b'TEST1=hello', b'TEST2=goodbye']

# Convert Python lists to c_char_p arrays.
# Arrays should be null-terminated so adds None to the end.
cargs = (ct.c_char_p * (len(args) + 2))(cmd, *args, None)
cenv = (ct.c_char_p * (len(env) + 1))(*env, None)

dll._execve(cmd, cargs, cenv)

输出：

argv[0] = "test.exe"
argv[1] = "arg1"
argv[2] = "arg2"
argv[3] = "arg3"
TEST1=hello
TEST2=goodbye