C 编程 - 在 main() 之外的函数中收集数据

Question

编写一个程序，要求用户输入每日降雨量。 您的程序需要接受 5 个每日降雨量输入。 只允许非负降雨量。 当用户输入负数时，告诉他们该数字无效，他们应该输入另一个有效值。

计算总降雨量和平均降雨量。 确定最大日降雨量和最小日降雨量。

使用信息性消息输出总数、平均值、最大值和最小值。

以下事情不能在 main 中发生：

1. 接受用户输入
2. 计算总数或平均值
3. 确定最大或最小
4. 输出结果

==============================================

现在我只是想弄清楚如何输入 5 个数字，到目前为止我已经有了这个代码，但它让我把它输入了无数次。 我已经在这个项目上工作了几个小时，所以任何建议都会很棒。

#include <stdio.h>
#include <stdlib.h>
#define SIZE 5 // have the user enter it 5 times

double CollectRainfall() {
    double amount;
    double rainfall[SIZE];
    int i;

    printf("Enter a rainfall amount: \n");  // enter amount
    scanf_s("%lf", &amount);

    for (i = 0; i < SIZE; i++) {
        rainfall[i] = CollectRainfall();
        while (amount < 0.0) {  // if it's a negative number
            printf("The number is invalid.\n");  // display error message if a negative # was entered
            printf("Enter another rainfall amount: \n");
        }
    }

}   
int main() {

    CollectRainfall();

    return 0;
}  

Answer 1

您可以创建一个结构来存储数据并执行操作。

就像是：

#include <stdio.h>
#include <stdlib.h>
#define SIZE 5 // have the user enter it 5 times


typedef struct data {
    double rainfall[SIZE];
    double average;
    double min;
    double max;
} data_t;

static void collectRainfall(double rainfall[SIZE]) {
    for (int i = 0; i < SIZE; i++) {
        double amount;

        printf("Enter a rainfall amount: \n");  // enter amount
        scanf("%lf", &amount);
        while (amount < 0.0) {  // if it's a negative number
            printf("The number is invalid.\n");  // display error message if a negative # was entered
            printf("Enter a rainfall amount: \n");  // enter amount
            scanf("%lf", &amount);
        }
        rainfall[i] = amount;
    }
}

static void compute(data_t *data) {
     data->min = data->rainfall[0];
     data->max = data->rainfall[0];
     data->average = data->rainfall[0];

     for (int i = 1; i < SIZE; i++) {
         double rainfall = data->rainfall[i];
         if (rainfall > data->max) {
             data->max = rainfall;
         }
         if (rainfall < data->min) {
             data->min = rainfall;
         }
         data->average += rainfall;
     }
     data->average /= SIZE;
}

static void display(data_t *data) {
    printf("min %f, max %f, average %f\n",
            data->min, data->max, data->average);
}

int main() {
    data_t data;

    collectRainfall(data.rainfall);
    compute(&data);
    display(&data);

    return 0;
}

scanf在输入错误的情况下很痛苦，最好是读取一行然后解析它，检查strtod是否正常

static void collectRainfall(double rainfall[SIZE]) {
    for (int i = 0; i < SIZE; i++) {
        char str[32];
        double amount = -1;

        printf("Enter a rainfall amount [%d/%d]: \n", i , SIZE);  
        while (42) {
            char *res = fgets(str, sizeof(str), stdin);
            if (res && (amount = strtod(str, &res)) >= 0 && res != str)
                break;
            printf("The number is invalid.\n");
            printf("Enter a rainfall amount [%d/%d]: \n", i , SIZE);  
        }
        rainfall[i] = amount;
    }
}

Answer 2

如前所述，递归实际上将创建一个无限循环，实际上，为此您也不需要它，您可以执行以下操作：

运行示例（注释更改）

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

#define SIZE 5 // have the user enter it 5 times

void CollectRainfall() { //no return needed

    double rainfall[SIZE], sum = 0, max = 0, min = 0;
    int i;


    for (i = 0; i < SIZE; i++)
    {
        printf("Enter a rainfall amount: \n"); // enter amount
        scanf("%lf", &rainfall[i]); //save values into the array
        while (rainfall[i] < 0.0)
        {                                       // if it's a negative number
            printf("The number is invalid.\n"); // display error message if a negative # was entered
            printf("Enter another rainfall amount: \n");
            i--; // iterate back to replace negative number
        }
    }
    printf("Values:");
    for (i = 0, min = rainfall[i]; i < SIZE; i++)
    {
        printf(" %.2lf", rainfall[i]); // print all values
        sum += rainfall[i];          // sum values
        if(rainfall[i] > max){       //max value
            max = rainfall[i];      
        }
        if(min > rainfall[i]){       //min value
            min = rainfall[i];
        }
    }
    printf("\nSum: %.2lf", sum);            // print sum
    printf("\nMax: %.2lf", max);            // print max
    printf("\nMin: %.2lf", min);            // print min
    printf("\nAverage: %.2lf", sum / SIZE); //print average
}

int main() {
    CollectRainfall();
    return 0;
}