c - linux如何修补这段代码

Question

#include <WhatHere?>
#include <WhatHere?>
#include <WhatHere?>
int main(int argc, char **argv) {
    char command[50] = "echo ";
    strcat(command,argv[1]); // concatenate the input so that the final command is "echo <input>"
    system(command); // call the system() function to print the input
    return 0; // denote that the program has finished executing successfully
}

我们可以通过运行这段代码来获得远程访问吗？我知道这是可能的，但请帮我修补它。

Answer 1

假设您担心潜在的缓冲区溢出，您可以这样修复它:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main (int argc, char **argv) {
    char *command;
    if (argc != 2) {
        fprintf (stderr, "Wrong number of arguments\n");
        return 1;
    }
    if ((command = malloc (strlen (argv[1]) + 6)) == NULL) {
        fprintf (stderr, "Could not allocate memory\n");
        return 1;
    }
    strcpy (command, "echo ");
    strcat(command,argv[1]);
    system(command);
    free (command);
    return 0;
}

这为 "echo " (5)、argv[1](字符串长度)和空终止符 (1) 留出了足够的空间。

允许运行用户指定的内容仍然存在潜在危险，但至少您不会再遇到缓冲区溢出。