c - 我的程序不打印最后一行代码的函数结果

Question

此代码接受用户的输入(字符 C、T、B)和(整数 0-24 和 0-60)，以根据用户输入的车辆类型计算 parking 费用。

程序中的最后一行代码应该打印函数“charged”的结果，该结果由用户输入的字符值声明的车辆类型决定，但当我运行时，它仅返回 0.00 而不是 flaot重视任何和所有的帮助:)

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



int total_minute_parked (int minute_in, int minute_left)
{
    int minute_parked;
    if (minute_in > minute_left)
    {
        minute_parked = (minute_left  - minute_in + 60);
    }
    else
    {
        minute_parked = (minute_left - minute_in);
    }
    return minute_parked;
}

// func calc total hours parked

int total_hour_parked (int hour_in, int hour_left)
{
    int hours_parked;
    if (hour_left > hour_in)
    {

        hours_parked = abs((hour_left - 1) - hour_in);
    }
    else
    {
        hours_parked = abs(hour_left - hour_in);
    }

    return hours_parked ;
}

// -------------------funtion to calc charge based off type of vehicle------

float charged (char vehicle_type,int total_hour_parked)
{

    char C;
    char T;
    char B;

    float temp_charged;

    if  (vehicle_type == C) // -------------------------------CAR
    {
        if (total_hour_parked > 3)
        {
            float secondary_hour = total_hour_parked - 3;
            temp_charged = secondary_hour * 1.5;
        }
        else
        {
            temp_charged = 0;
        }
    }

    else if  (vehicle_type == T)   // ------------------------------TRUCK
    {
        if (total_hour_parked > 2)
        {
            float secondary_hour = total_hour_parked - 2;
            temp_charged = (secondary_hour * 2.3) + 1.0;
        }
        else {
            temp_charged = 1;
        }
    }

    else if  (vehicle_type == B) // -----------------------------------BUS
    {
        if (total_hour_parked > 1)
        {
            float secondary_hour = total_hour_parked - 1;
            temp_charged = (secondary_hour * 3.7) + 2.0;
        }
        else {
            temp_charged = 2;
        }
    }
    return temp_charged;
}

//---------------------- end program upon invalid imput -------------------//



// --------------------- main that prints results and takes imput -----------//

int main()

{

    int total_hour_parked (int hour_in,int hour_left);
    float charged (char vehicle_type, int total_hour_parked);
    char vehicle_type;
    int hour_in = 0;
    int minute_in = 0;

    int hour_left = 0;
    int minute_left = 0;

    printf("Please enter the type of Vehicle:");
    scanf("%c",&vehicle_type);

    printf("Please enter the hour entered lot:");
    scanf("%d", &hour_in);

    printf("Please enter the minute entered lot:");
    scanf("%d", &minute_in);

    printf("Please enter the hour left lot:");
    scanf("%d", &hour_left);

    printf("Please enter the minute left lot:");
    scanf("%d", &minute_left);

    printf("------------------------------------\n");

    printf("Parking time: %d:%d\n", total_hour_parked(hour_in,hour_left),total_minute_parked(minute_in,minute_left));

    printf("Cost %f",charged(vehicle_type,total_hour_parked));

    return 0;
}

Answer 1

我不确定它是否能解决代码中的所有问题，但这里有一个问题:

char C;
char T;
char B;

float temp_charged;

if  (vehicle_type == C) // -------------------------------CAR 

它的作用是声明三个char并且不分配任何值(因此访问它们是未定义的并且会导致一些垃圾值)。然后，您将 vehicle_type 与这些字符进行比较。结果很可能是false(或0)。那不是你的本意。相反，请执行以下操作:

if  (vehicle_type == 'C')

您可能误解了 char C; 的含义。它并不意味着“创建一个值为 'C' 的 char”，而是“创建一个名为 C 的 char > 并且不初始化它”。但在这种情况下，您无论如何都不需要这三个字符，因为您只需将 vehicle_type 与文字进行比较即可。