字符串问题（do/time循环和if逻辑）

Question

这段代码在两个方面出了问题，我无法诊断，但与if语句无关:if (strlen(strA)

我注意到的第二个问题是，如果我输入一个字符串“1111111111111”，它显然太长，程序会说密码太长，但整个do/while循环终止，它会要求我确认密码。它应该要求我再次输入密码。

如果我去掉 if 语句：if （strlen（strA）

任何人都可以诊断问题吗？

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

int main(void) {
    char strA[10];
    char strB[10];
    char strC[] = {'1','2','3','4','5','6','7','8','9','0'};
    char strD[] = {'!','@','#','$','%','^','&','*','(',')'};

    char strE[] = {'a','b','c','d','e','f','g','h','i','j','k',
                   'l','m','n','o','p','q','r','s','t','u','v',
                   'w','x','y','z'};

    char strF[] = {'A','B','C','D','E','F','G','H','I','J',
                   'K','L','M','N','O','P','Q','R','S',
                   'T','U','V','W','X','Y','Z'};
    int i, j, k;
    do {
        k = 0;
        j = 0;

        printf("Please enter your password: ");
        scanf("%s", &strA);
        printf("%s\n", strA);

       if(strlen(strA) > 8) {
           printf("That password is too long\n");
       }
       else if(strlen(strA) < 9) {
           while (j <= 9) {
               for(i = 0; i <= 9; i++) {
                   if(strA[j] == strC[i]) {
                       printf("there is a number in this string.\n");
                       k++;
                       j = 0;
                       while (j <= 9) {
                           for(i = 0; i <= 9; i++) {
                               if(strA[j] == strD[i]) {
                                   printf("there is a character in this string.\n");
                                   k++;
                                   j = 0;
                                   while(j <= 9) {
                                       for(i = 0; i <= 25; i++) {
                                           if(strA[j] == strE[i]) {
                                               printf("there is a lowercase letter in this string.\n");
                                               k++;
                                               j = 0;
                                               while(j <= 9) {
                                                   for(i=0;i<=25;i++) {
                                                       if(strA[j] == strF[i]) {
                                                           printf("there is an uppercase letter in this string.\n");
                                                           k++;
                                                       }
                                                   }
                                                   j++;
                                              }
                                          }
                                      }
                                      j++;
                                  }
                              }
                          }
                          j++;
                      }
                  }
              }
              j++;
          }
          if(k < 4) {
              printf("Your password must contain at least one uppercase letter, one lowercase letter, a number, and a special character.\n");
          }
      }
    } while(k < 4);

    printf("Please confirm your password: ");
    scanf("%s",&strB);

    while(strcmp(strA, strB) != 0) {
        printf("%s\n",strB);
        printf("Your passwords do not match.\nPlease confirm your password: ");
        scanf("%s",&strB);
    }

    putchar('\n');
    printf("%s\n", strA);
    printf("%s\n", strB);

    return 0;
}

Answer 1

查看ASCII标准：有一种更快的方法可以测试char c是否是数字（c

更多信息:查看此处所述的< code>ctype.h,确定char是数字还是字母

http://www.cplusplus.com/reference/locale/isalpha/

isalpha（c）：如果字母

isdigit（c）：如果数字为真

''isupper（c）' ： 如果大写，则为 true

< code>islower(c):如果小写，则为true

再见，

弗朗西斯

Answer 2

我认为下面这句话是在制造问题:

char strA[10];

char strB[10];

使用默认值初始化

memset(&strA,'\0', sizeof(strA));
memset(&strB,'\0', sizeof(strB));