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谁能帮我算一下矩形的面积。我不能做到万无一失，即禁止输入字母和负数。我只是不能在创建第一个涉及字母的条件时创建第二个条件。我该怎么做，或者我什至必须从头开始更改程序。我只需要找到一个矩形的面积。

#include "stdafx.h"
#include <stdio.h>
#include <math.h>
#include <locale.h>
#include <Windows.h>

int _tmain(int argc, char* argv[])
{
    printf("Area of a Rectangle. Press Enter to start\n");
    system("pause");

    float a, s, d;

    do
    {
        fflush(stdin);
        system("cls");
        printf("Enter the lengths of the rectangle\n");
        scanf_s("%f", &a);
        scanf_s("%f", &s);

        if (getchar() != '\n') {
            while (getchar() != '\n')
                ;
            printf("Your data is wrong\n");
            system("pause");
            continue;
        }

        break;
    } while (true);

    d = a*s;
    printf(" Area is %.1f ", d);
    system("pause");
    return 0;
}
4

2 回答 2

2

这个程序将要求一个数值并且只接受正值。输入字母或符号将导致scanf()失败，程序将清除输入缓冲区并重试。在本例中，输入字母“q”将退出程序。您可以根据自己的情况进行调整。

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


int main()
{
    int i = 0;
    float n = 0.0f;

    do {
        printf("enter numeric value or q to quit\n");
        if ( scanf("%f",&n) != 1) {// scan one float
            while ( ( i = getchar ( )) != '\n' && i != 'q') {
                //clear buffer on scanf failure
                //stop on newline
                //quit if a q is found
            }
            if ( i != 'q') {
                printf ( "problem with input, try again\n");
            }
        }
        else {//scanf success
            if ( n == fabs ( n)) {
                printf("number was %f\n", n);
            }
            else {
                printf("positive numbers only please\n");
            }
        }
    } while ( i != 'q');

    return 0;
}

这已将上述内容改编为一个函数。

#include <stdio.h>

float getfloat ( char *prompt, int *result);

int main ( int argc, char* argv[])
{
    float width, height, area;
    int ok = 0;

    do {
        printf("\n\tArea of a Rectangle.\n");
        width = getfloat ( "Enter the width of the rectangle or q to quit\n", &ok);
        if ( ok == -1) {
            break;
        }
        height = getfloat ( "Enter the height of the rectangle or q to quit\n", &ok);
        if ( ok == -1) {
            break;
        }

        area = width * height;
        printf(" Area is %.1f\n", area);
    } while (1);

    return 0;
}

float getfloat ( char *prompt, int *result)
{
    int i = 0;
    float n = 0.0f;

    *result = 0;

    do {
        printf("%s", prompt);
        if ( scanf("%f",&n) != 1) {// scan one float
            while ( ( i = getchar ( )) != '\n' && i != 'q') {
                //clear buffer on scanf failure
                //stop on newline
                //quit if a q is found
            }
            if ( i != 'q') {
                printf ( "problem with input, try again\n");
            }
            else {
                *result = -1;
                n = 0.0f;
            }
        }
        else {//scanf success
            if ( n == fabs ( n)) {
                *result = 1;
            }
            else {
                printf("positive numbers only please\n");
            }
        }
    } while ( *result == 0);

    return n;
}
于 2015-09-07T19:59:11.790 回答
1

浮动的万无一失scanf看起来像这样

float getFloat() {
    float in;
    while(scanf("%f", &in) != 1 && getchar() != '\n') {
        fflush(stdin);
        fprintf(stderr, "Wrong input, enter again : ");
    }
    return in;
}

这对我有用。然后您的代码将简化为

float a = getFloat();
float b = getFloat();
printf("Area is %.1f", a * b);
于 2015-09-08T12:52:17.140 回答