Question

我正在尝试调用是/否函数来查看用户是否输入了有效输入，在这种情况下是'Y''''N''n'。但我不希望输出之间有两条新线，你可以从下面的附图中看到。

Incorrect Part

clearKeyboard功能： void clearKeyboard（void）

{
    int c;
    while ((c = getchar()) != '\n' && c != EOF);
}

是/否功能：

int yes(void)
{
    int result, done = 1;
    char c = 0, charenter;

    scanf("%c%c", &charenter, &c);

    while (done == 1) {
        if (c != '\n' || (!(charenter == 'Y' || charenter == 'y' || charenter == 'N' || charenter == 'n'))) {
            clearKeyboard();
            printf("*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>: ");
            scanf("%c%c", &charenter, &c);
        }
        else {
            done = 0;
        }
    }

    if (charenter == 'Y' || charenter == 'y')
        result = 1;
    else if (charenter == 'N' || charenter == 'n')
        result = 0;

    return result;
}

调用yes / no函数的函数：我试过删除fflush（stdin）;或者在％c之前留一个空格，没有一个工作。但“家庭电话号码”部分工作正常，没有给我新行/输入错误无效。

void getNumbers(struct Numbers *numbers) // getNumbers function definition
{
    char answer;
    int result;

    printf("Please enter the contact's cell phone number: ");
    scanf("%s", numbers->cell);

    printf("Do you want to enter a home phone number? (y or n): ");
    scanf("%c", &answer);
    result = yes();
    while (result == 1) {
        printf("Please enter the contact's home phone number: ");
        scanf("%s", numbers->home);
        break;
    }

    printf("Do you want to enter a business phone number? (y or n): ");
    fflush(stdin);
    scanf("%c", &answer);
    result = yes();
    while (result == 1) {
        printf("Please enter the contact's business phone number: ");
        scanf("%s", numbers->business);
        break;
    }
}

提前致谢！

Answer 1

我建议使用fgets代替scanf，您可以更好地控制使用fgets的行。

如果您使用scanf来读取字符串，我会使用此函数来清除缓冲液：

void clear_buffer(FILE *fp)
{
    int c;
    while((c=fgetc(fp)) != '\n' && c!=EOF);
}

然后你可以做

char name[15];
char phone[15];
scanf("%14s", name);
clear_buffer(stdin);

scanf("%14s", phone);
clear_buffer(stdin);

因此，您的yes函数可以重写为：

int yes(const char *prompt)
{
    char line[10];
    int ret;

    do {
        printf("%s: ", prompt);
        fflush(stdout);

        ret = scanf("%9s", line);
        clear_buffer(stdin);

        if(ret == 1)
        {
            // catch entries like "yes" and "nooooooo"
            if(line[1] != 0)
                line[0] = 0;

            switch(line[0])
            {
                case 'y':
                case 'Y':
                    return 1;

                case 'n':
                case 'N':
                    return 0;

                default:
                    printf("*** INVALID ENTRY *** <Only (Y)es or (N)o are acceptable>\n");
            }
        }
    } while(ret != EOF);

    return 0;
}

然后你可以读出这样的数字：

void getNumbers(struct Numbers *numbers) // getNumbers function definition
{
    int result;

    printf("Please enter the contact's cell phone number: ");
    scanf("%s", numbers->cell);
    clear_buffer(stdin);


    if(yes("Do you want to enter a home phone number? (y or n)"))
    {
        printf("Please enter the contact's home phone number: ");
        scanf("%s", numbers->home);
        clear_buffer(stdin);
    }

    if(yes("Do you want to enter a business phone number? (y or n)"))
    {
        printf("Please enter the contact's business phone number: ");
        scanf("%s", numbers->business);
        clear_buffer(stdin);
    }
}

顺便问一下：你为什么要这样做

while (result == 1) {
    printf("Please enter the contact's home phone number: ");
    scanf("%s", numbers->home);
    break;
}

而不是

if (result == 1) {
    printf("Please enter the contact's home phone number: ");
    scanf("%s", numbers->home);
}

产生相同的结果，但这更容易阅读。

调用（是/否）函数以确定用户是否输入有效输入

1 个答案: