当字符和整数混合时控制 C 中的用户输入

Question

所以我正在尝试为只接受选项 1,2 和 3 的程序制作一个简单的菜单，我想控制用户输入以防止错误，例如当用户输入 char 而不是 int 时。

现在它控制一些情况，比如如果选项小于 1 或大于 3，如果选项是一个字符它也不会影响但是当用户输入类似“02”或“2a”的东西时它运行选项 2 但应该使该选项无效。

此外，如果我遗漏了更多案例，我想了解它们以及如何克服它们。

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

void empty_stdin(void);

int main() {
    int option;
    int rtn;

    do {
        printf("\n--\nOptions:\n1.Option 1\n2.Option 2\n3.Option 3\n--\n\nPlease chose option (1/2/3) to continue: ");
        rtn = scanf("%d", &option);

        if (rtn == 0 || option < 1 || option > 3) {
            printf("-Invalid Option-\n");
            empty_stdin();
        } else {

            empty_stdin();

            switch (option) {
                case 1:
                    printf("Option 1");
                    break;

                case 2:
                    printf("Option 2");
                    break;
                case 3:
                    printf("Option 3");
                    exit(0);

                default:
                    printf("\n-Invalid Option-\n");
            }
        }
    } while (option != 3);
    return 0;
}

void empty_stdin(void) {
    int c = getchar();

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

示例Input/Output

--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 1
Option 1
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 12
-Invalid Option-

--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: char
-Invalid Option-

--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 02
Option 2
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 2a
Option 2
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 02a
Option 2
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

预计Input/Output

--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 1
Option 1
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 12
-Invalid Option-

--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: char
-Invalid Option-

--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 02
-Invalid Option-
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 2a
-Invalid Option-
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Please chose option (1/2/3) to continue: 02a
-Invalid Option-
--
Options:
1.Option 1
2.Option 2
3.Option 3
--

Answer 1

考虑使用 fgets 捕获输入并使用 strtol 解析整数。

#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <limits.h>

int fgetsint ( int *value, int min, int max, FILE *fin) {
    char line[100] = "";
    char extra = '[=10=]';
    char *end = NULL;
    long int number = 0;

    if ( fgets ( line, sizeof line, fin)) {//read a line
        errno = 0;
        number = strtol ( line, &end, 10);
        if ( end == line) {// nothing was parsed. no digits
            printf ( "input [%s] MUST be a number\n", line);
            return 0;// return failure
        }
        if ( ( errno == ERANGE && ( number == LONG_MAX || number == LONG_MIN))
        || ( errno != 0 && number == 0)) {// parsing error from strtol
            perror ( "input error");
            return 0;
        }
        if ( 1 == sscanf ( end, " %c", &extra)) {//parse trailing character
            printf ( "enter one number only. try again\n");
            return 0;
        }
        if ( number > max || number < min) {
            printf ( "Input [%ld] out of range: min: %d max: %d\n", number, min, max);
            return 0;
        }
        if ( 1 != (int)( end - line)) {
            printf ( "input one digit\n");
            return 0;// return failure
        }
        *value = number;//assign number to pointer
    }
    else {
        fprintf ( stderr, "problem fgets\n");
        exit ( EXIT_FAILURE);
    }

    return 1;//success
}

int main ( void) {
    int option = 0;

    do {
        printf("\n--\nOptions:\n1.Option 1\n2.Option 2\n3.Option 3\n--\n\nPlease chose option (1/2/3) to continue: ");

        option = 0;
        fgetsint ( &option, 1, 3, stdin);

        switch (option) {
            case 1:
                printf("Option 1");
                break;

            case 2:
                printf("Option 2");
                break;
            case 3:
                printf("Option 3");
                exit(0);

            default:
                printf("\n-Invalid Option-\n");
        }
    } while ( option != 3);
    return 0;
}

Answer 2

建议代码如下：

干净地编译
执行所需的功能
处理了所有 'oops' 个案例
尊重右边距，用于打印等

现在，建议的代码：

#include <stdio.h>

void empty_stdin( void );

int main( void ) 
{
    int option = 9;

    do 
    {
        printf("\n--\nOptions:\n"
               "1.Option 1\n"
               "2.Option 2\n"
               "3.Option 3\n"
               "--\n\n"
               "Please chose option (1/2/3) to continue: ");
        option = getchar();

        empty_stdin();

        switch (option) 
        {
            case 1:
                printf( "Option 1\n" );
                break;

            case 2:
                printf( "Option 2\n" );
                break;

            case 3:
                printf( "Option 3, exiting\n" );
                //exit(0);
                break;

            default:
                printf( "\n-option: %d invalid-\n", option );
                break;
        }
    } while ( option != 3 && option != EOF );
    return 0;
}

void empty_stdin( void ) 
{
    //int c = getchar();

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

当字符和整数混合时控制 C 中的用户输入

Control user input in C when there is chars and ints mixed

c

defensive-programming

input

scanf