Question

我正在编写一个代码，要求用户输入10个整数，然后将这些整数反馈给他。我想创建一个＆＃34; scanf check＆＃34;限制字符输入。 while循环可以在不接受char的情况下工作，但它会跳过整数输入。

int main()
{

    int i = 0, number[10] = {0};
    char buf[128] = {0};

    for (i = 0; i < 10; i++)
    {
      printf("Please input number %d : ", i+1);

         while(scanf("%d", &number[i]) != 1)
      {
         scanf("%s", &buf);
         printf("Sorry, [%s] is not a number. Please input number %d : ", &buf, i);
      }
    }

    for (i = 0; i < 10; i++)
    {
    printf("\n Number %d is %d", (10-i), number[9-i]);
    }

    return EXIT_SUCCESS;
}

Answer 1

正如H2CO3所指出的，不要使用scanf，另一种选择是fgets和strtol：

int i, number[10] = {0};
char buf[128], *p;

for (i = 0; i < 10; i++) {
    printf("Please input number %d : ", i+1);
    while (1) {
        fgets(buf, sizeof(buf), stdin);
        if ((p = strchr(buf, '\n')) != NULL) {
            *p = '\0';
        }
        number[i] = (int)strtol(buf, &p, 10);
        if (p == buf || *p != '\0')  {
            printf("Sorry, [%s] is not a number. Please input number %d : ", buf, i + 1);
        } else {
            break;
        }  
    }
}
for (i = 0; i < 10; i++) {
    printf("\n Number %d is %d", (10-i), number[9-i]);
}
return EXIT_SUCCESS;

Answer 2

代码也适用于整数。我发现的唯一错误是在打印抱歉邮件时，您只打印i，它应该是i+1。

    int i = 0, number[10] = {0};
    char buf[128] = {0};

    for (i = 0; i < 10; i++)
    {
      printf("Please input number %d : ", i+1);

         while(scanf("%d", &number[i]) != 1)
      {
         scanf("%s", &buf);
         printf("Sorry, [%s] is not a number. Please input number %d : ", &buf, i+1);
      }
    }

    for (i = 0; i < 10; i++)
    {
    printf("\n Number %d is %d", (10-i), number[9-i]);
    }

scanf check in while循环以限制整数输入

2 个答案: