c - 卡在 C 中的赋值上

Question

大家好，我是编程新手，对于我的编程类(class)介绍，我在完成这项作业时遇到了一些问题。我需要一些帮助来解决这个问题，即创建一个输出总降雨量、平均降雨量、最大降雨量和最小降雨量的程序。一切都很好，直到输出最小的降雨量，它一直输出为 0.0。

#include <stdio.h>
#include <stdlib.h>
#define SIZE 5

double getValidRainfall(){
    double result;

    printf("Enter a rainfall amount: ");
    scanf("%lf", &result);

    while (result < 0.0){
        printf("%.2lf is not a valid rainfall amount.", result);
        printf("\nEnter a rainfall amount: ");
        scanf("%lf", &result);
    }

    return result;
}

double calculateTotal (double rainfall[], int size){
    double result = 0.0;
    int i;
    for(i = 0; i < size; i++){
        result += rainfall[i];
    }
    return result;
}

double findLargest(double rainfall[], int size){
    double result = 0.0;
    int i;
    for (i = 0; i < size; i++) {
        if(rainfall[i] > result)
            result = rainfall[i];
    }
    return result;
}

// I believe the problem is located right under here and...

double findSmallest(double rainfall[], int size){
    double result = 0.0;
    int i;
    for (i = 0; i < size; i++) {
        if(rainfall[i] < result)
            result = rainfall[i];
    }
    return result;
}

    // here.. Just can't seem to find the issue that will output the correct smallest
    // amount of rainfall.

int main(){
    double rainfall[SIZE], totalRainfall, averageRainfall;
    double largestRainfall, smallestRainfall;
    int i;

    for( i = 0; i < SIZE; i++)
        rainfall[i] = getValidRainfall();


    totalRainfall = calculateTotal(rainfall, SIZE);
    averageRainfall = totalRainfall / SIZE;
    largestRainfall = findLargest(rainfall, SIZE);
    smallestRainfall = findSmallest(rainfall, SIZE);

    printf("Thank you\n");
    printf("The total rainfall is %.2lf\n", totalRainfall);
    printf("The average rainfall is %.2lf\n", averageRainfall);
    printf("The largest daily rainfall is %.1lf\n", largestRainfall);
    printf("The smallest daily rainfall is %.1lf\n", smallestRainfall);
}

Answer 1

在 findSmallest 中，您从:

double result = 0.0;

然后你只在降雨量较小时更新它。然而，假设没有任何地方的降雨量为负值，这意味着最终结果仍为 0.0。

要使该算法起作用，您应该以至少与任何实际降雨量一样高的值开始，例如

 if ( size == 0 )
     return 0.0;

 double result = rainfall[0];