Question

我正在为字母/数字ASCII表创建一个转换项目。我的代码应该是“交互式的”，因此用户可以在屏幕上输入“ y”或“ n”来回答问题。但是，它不想重复两次...

我尝试过：

只尝试数字而不是字符，但这不是我想要的
%[\n]*c，%[\n]c和%[\n]*s ...技术，但无济于事；-;
在不同的项目中进行测试，但我唯一能做的是将多个scanf()连续放置。

代码如下：

printf("Would you like to convert a number today? \n");
printf("Please press Y or N \n");
scanf("%c", &input);

if (input == 'y' || input == 'Y') { //compare input if they said 'yes'
    printf("\nThank you! \nWhat number?\n");
    scanf("%d", &number);
    flag = test(number);
    if (flag == 0) { //if there is an equivalent letter
        letter = conversion(number); //find the equivalent letter
        printf("\nYour Number \t ASCII letter\n");
        printf("%d\t %c\n", number, letter);
    }
}
else if (input == 'n' || input == 'N') {
    printf("\nWould you like to convert a letter instead? This time enter 0 or 1\!\n\n"); //problem here!!
    printf("I wish I could say it was to \' Spice things up \' ...but it\'s not ;-; \n\n");
    scanf("%d", &input2);

    if (input2 == 0) { //this needs to be checking whether the user input Y/y
        printf("Great choice adventurer!\n");
        printf("What letter will it be today?\n\n");

        //..I would go to a different funtion here ie: test2(letter)...

        scanf("%d", &number); //I showed that it worked with multiple numbers, but I can't get this to work with multiple letters
        printf("%d", number);
    }

    if (input2 == 1) { //this needs to be checking whether the user input N/n
        printf("Difficult to please, I see...\n\n");
        printf("I suggest you move on with that attitude!\n\n");
        printf("Bye bye then\n");
    }
}
else { //if they tried to break the code
    printf("Sorry I did not recognise your command...please retry\n");
    printf("Press Y or N next time!\n");
}

第一张支票可以完美工作，我只希望第二张支票像第一张支票一样！一些“解决方案”导致了溢出，如果可能的话，我不希望这样做即使有人可以解释为什么这没有按照我的意图工作也会很有帮助！

Answer 1

我不确定是什么让您感到困惑。

使用

char foo;
scanf(" %c", &foo);

用于单个字符，例如。字母和

int bar;
scanf("%d", &bar);

用于数字，整数。如果您键入字母，则scanf()将失败。

%[...]用于字符串。

scanf()返回成功的转换次数（或EOF），因此

int height;
int width;
scanf("%d %d", &height, &width);

如果成功，它将返回2。如果只能读取1，则可能返回height。

因此，要检查用户输入是否有错误，您应该执行以下操作：

int height;
int width;
if (scanf("%d %d", &height, &width) != 2) {
    // handle the error, maybe exit the program.
}

您的代码可能看起来像这样（没有错误处理）：

#define _CRT_SECURE_NO_WARNINGS  // you said Visual Studio? Without it you should get
                                 // warnings about some functions being insecure.

#include <ctype.h>   // isalpha()  returns true if the value is a letter
#include <stdlib.h>  // EXIT_SUCCESS
#include <stdio.h>   // puts(), printf(), scanf()

int main(void)
{
    for(;;) {  // for-ever ... endless loop since the user exits by answering
               // 'n' or 'N' two times

        puts("Would you like to convert a number today?\nPlease press Y or N:");
        char input;
        if (scanf(" %c", &input) != 1)  // We reached EOF ... end of file
            break;                      // that's improbable for stdin,
                                        // but input could be redirected to
                                        // read from a file instead.

        if (input == 'y' || input == 'Y') {
            puts("\nThank you!\nWhat number?");
            int number;
            scanf("%d", &number);

            if (isalpha((char unsigned)number))  // *)
                printf("\nYour Number \t ASCII letter\n%d\t %c\n\n", number, number);
            else
                puts("Sorry, but that's not the ASCII code of a letter :(\n");
        }
        else if (input == 'n' || input == 'N') {
            puts("\nWould you like to convert a letter instead?\nPlease press Y or N:");
            scanf(" %c", &input);

            if (input == 'y' || input == 'Y') {
                puts("\nGreat choice adventurer!\nWhat letter will it be today?");
                char letter;
                scanf(" %c", &letter);

                if (isalpha(letter))
                    printf("\nYour letter \t ASCII code\n%d\t %c\n\n", letter, letter);
                else
                    puts("Sorry, but that's not a letter :(\n");
            }
            else if (input == 'n' || input == 'N') {
                puts("\nDifficult to please, I see...\n\nI suggest you move on with that attitude!\n");
                puts("Bye bye then.");
                return EXIT_SUCCESS;
            }
        }
        else {
            puts("Sorry I did not recognize your command... Please retry.");
            puts("Press Y or N next time!\n");
        }
    }
}

*）isalpha()（以及<ctype.h>中的其他函数）期望的值适合unsigned char或值EOF。对于其他值，它具有未定义的行为。由于我们将用户输入读入int中，因此我们无法确定是否是这种情况，因此我们必须在将值传递给unsigned char（和朋友）之前将其转换为isalpha()。

下次您提出问题时，请提供完整的代码，包括变量声明，函数test()和conversion()和#include。但是，请发布一个示例，以解决您手头的问题。您包括的所有对话框都是不必要的。

Visual Studio中的Scanf不接受多个字符的情况

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