c ++将argv读入unsigned char固定大小:分段错误

c++ reading argv into unsigned char fixed size: Segmentation fault

我正在尝试将命令行参数读入固定大小的无符号字符数组。我遇到分段错误。

我的代码:

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

unsigned char key[16]={};

int main(int argc, char** argv){
        std::cout << "Hello!" << std::endl;
        long a = atol(argv[1]);
        std::cout << a << std::endl;
        memcpy(key, (unsigned char*) a, sizeof key);
//      std::cout << sizeof key << std::endl;
//      for (int i = 0; i < 16; i++)
//              std::cout << (int) (key[i]) << std::endl;
        return 0;
}

我做错了什么?

调用程序:

编译:g++ main.cpp

执行:./a.out 128

您收到 SEGV 是因为您的地址错误:您将值转换为地址。加上大小是目标之一,应该是源的大小

编译器发出警告,这从来都不是好事,您应该考虑到这一点,因为那正是您的错误:

xxx.c:12:38: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]

     memcpy(key, (unsigned char*) a, sizeof key);
                                  ^

像这样修复:

memcpy(key, &a, sizeof(a));

顺便说一句,您不必用 16 个字节声明 key。像这样分配它会更安全:

unsigned char key[sizeof(long)];

并且当您打印字节时,也要迭代直到 sizeof(long),否则您最后只会打印垃圾字节。

这是一个使用 uint64_t 的修复建议(来自 stdint.h 的无符号 64 位整数,它可以精确控制大小),key 的零初始化和使用 strtoll:

#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <memory.h>
#include <stdint.h>

unsigned char key[sizeof(uint64_t)]={0};

int main(int argc, char** argv){
        std::cout << "Hello!" << std::endl;
        uint64_t a = strtoll(argv[1],NULL,10);
        memcpy(key, &a, sizeof a);

      for (int i = 0; i < sizeof(key); i++)
              std::cout << (int) (key[i]) << std::endl;
        return 0;
}

(如果要处理signed,改成int64_t即可)

小端架构测试:

% a 10000000000000
Hello!
0
160
114
78
24
9
0
0

您似乎复制了太多数据。 我还为 memcpy 添加了 &a。

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

unsigned char key[16]={};

int main(int argc, char** argv)
{
   memset(key,0x0, sizeof(key));
   std::cout << "Hello!" << std::endl;
   long a = atol(argv[1]);
   std::cout << a << std::endl;

   // the size parameter needs to be the size of a
   // or the lesser of the size of key and a
   memcpy(key,(void *) &a, sizeof(a));
   std::cout << "size of key " << sizeof(key) << "\n";
   std::cout << "key " << key << "\n";
   for (int i = 0; i < 16; i++)
   std::cout << "   " << i << "    '"  << ((int) key[i]) << "'\n";
   return 0;
}