strncpy segfault

Question

我一直无法使这段代码工作.我正在尝试获取要复制的字符数组,因此我可以计算出有多少令牌可以动态分配并保存它们以便检查环境变量.但是,当它试图strncpy原始字符串时,我保持segfaulting.

    void echo(char *str1)
    {
      char *token, *temp;
      char *saveptr1;
      int j, i, k, counter;
      char *copy;

      strncpy(copy, str1, 80);

      const char *delim = " ";
      i = strlen(copy);

      for(j = 0; j < i; j++, copy = NULL)
      {
         token = strtok_r(copy, delim, &saveptr1);
         counter++;
         if(token == NULL)
         {
           counter--;
           break;
         }
      }

      // initialize token array for echo
      char *tokAr[counter];
      for(j = 0; j < counter; j++)
        tokAr[j] = malloc(80*sizeof(char));

      for(j = 0, k = 0; j < i; j++, str1 = NULL)
      {
         tokAr[k] = strtok_r(str1, delim, &saveptr1);
         if( tokAr[k] != NULL)
         {
            if(strchr(tokAr[k], 36) != NULL)
            {
              temp = enviro(tokAr[k]);
              printf("%s ", temp);
            }
         else
           printf("%s ", tokAr[k]);
         }
         else
           break;
      }

      for(k = 0; k < counter; k++)
        free(tokAr[k]);
    }

    char* enviro(char *ret)
    {
      char *copy, *expand, *saveptr;
      const char *delim = "$";
      strcpy(copy, ret);
      expand = strtok_r(copy, delim, &saveptr);

      return getenv(expand);
    }

我知道它与我如何复制传入的str1字符数组有关,但我无法从gdb中找出它.任何帮助是极大的赞赏

Answer 1

你还没有分配内存copy.

char *copy;
strncpy(copy, str1, 80);

尝试malloc或strdup如果您不需要完整的81个字符.

copy = malloc(81);
strncpy(copy, str1, 80);

/* Or strdup. */
copy = strdup(str1);