XOR 来自 C 中 argv 的两个字符串

Question

想要对从 argv 中获取的两个字符串进行异或运算。我检查了这个问题但它无法为我解决。

#include <stdio.h>
#include <string.h>

int main(int argc, char const *argv[]) {
  char output[]="";

  int i;
  for (i=0; i<strlen(argv[1]); i++){
      char temp = argv[1][i]^argv[2][i];
      output[i]=  temp;

  }
  output[i] = '[=10=]';
  printf("XOR: %s\n",output);
  return 0;
}

当我使用 lldb 调试我的输出（“（lldb）打印输出”）时，它是 /a/x16/t/x13 但是printf()不能打印出来。我知道它不再是字符串了。你能帮我如何让它成为 printf:ed 吗？终端中打印的文本是 "XOR: "

Answer 1

您的代码中存在一些内存错误。也许以下会更好：

#include <stdio.h>
#include <string.h>

#define min(i, j) ((i) < (j) ? (i) : (j))

int main(int argc, char const *argv[])
  {
  char *output;
  int i;

  /* Allocate a buffer large enough to hold the smallest of the two strings
   * passed in, plus one byte for the trailing NUL required at the end of
   * all strings. 
   */

  output = malloc(min(strlen(argv[1]), strlen(argv[2])) + 1);

  /* Iterate through the strings, XORing bytes from each string together
   * until the smallest string has been consumed. We can't go beyond the
   * length of the smallest string without potentially causing a memory
   * access error.
   */

  for(i = 0; i < min(strlen(argv[1]), strlen(argv[2])) ; i++)
      output[i] = argv[1][i] ^ argv[2][i];

  /* Add a NUL character on the end of the generated string. This could
   * equally well be written as
   *
   *   output[min(strlen(argv[1]), strlen(argv[2]))] = 0;
   *
   * to demonstrate the intent of the code.
   */

  output[i] = '[=10=]';

  /* Print the XORed string. Note that if characters in argv[1]
   * and argv[2] with matching indexes are the same the resultant byte
   * in the XORed result will be zero, which will terminate the string.
   */

  printf("XOR: %s\n", output);

  return 0;
  }

就 printf 而言，请记住 x ^ x = 0 并且 [=13=] 是 C 中的字符串终止符。

祝你好运。

XOR 来自 C 中 argv 的两个字符串

XOR two strings from argv in C

c

string

xor

argv