如果语句在 7 小时内出现问题 JAVA 闰年 [关闭]

Question

我进入 Java 课程 4 周，经验为 0。 我被要求创建一个程序，该程序要求输入月份的数字表示示例 1 将是一月，也是一年的输入，并让程序输出 显示月份 如何一个月中有很多天和如果是闰年。这是我缩短的许多尝试之一。我做错了什么，它从 1 月开始，输入无关紧要。

package leapmonth;

import javax.swing.JOptionPane;

public class LeapMonth {
    public static void main(String[] args) {

        int Year;
        int MonthNumber;

    String MonthString;    


    JOptionPane.showMessageDialog(null, " HELLO, WELCOME TO LEAPMONTH ");
    MonthString = JOptionPane.showInputDialog( " Please enter the numerical respresentation of the month");
    MonthNumber = Integer.parseInt(MonthString);
    MonthString = JOptionPane.showInputDialog( " Please enter the year");
    Year = Integer.parseInt(MonthString);

    if( MonthNumber == 1 && Year % 4 == 0 && Year % 100 != 0 || Year % 400 == 0)
    {
     JOptionPane.showMessageDialog(null,"The month is January. " + "January has 31 days in it" + " and is also a leap year "); 
    }   
    else if( MonthNumber == 1 && Year % 4 != 0 && Year % 100 == 0 || Year % 400 != 0)
    {
     JOptionPane.showMessageDialog(null,"The month is January. " + "January has 31 days in it" + " and it is not a leap year ");   
    }
    else if( MonthNumber == 2 && Year % 4 == 0 && Year % 100 != 0 || Year % 400 ==0)
    {
     JOptionPane.showMessageDialog(null,"The month is February. " + "February has 29 days in it because this is a leap year ");  
    } 
    else if( MonthNumber == 2 && Year % 4 != 0 && Year % 100 == 0 || Year % 400 != 0)
    {
    JOptionPane.showMessageDialog(null,"The month is February. " + "February has 28 days in it because this is not a leap year ");
    }
   }
}

Answer 1

这并不意味着你认为它意味着：

MonthNumber == 1 && Year % 4 == 0 && Year % 100 != 0 || Year % 400 == 0

我想你想要的是这样的：

MonthNumber == 1 && Year % 4 == 0 && (Year % 100 != 0 || Year % 400 == 0)

强调括号。

解释：|| 和 && 运算符具有相同的优先级，因此它们从左到右关联。

如果没有括号，你的逻辑是有缺陷的，因为任何年份是 400 的倍数都会产生 true。

例如，假设Year=2000 和MonthNumber=25

然后

MonthNumber == 1 && Year % 4 == 0 && Year % 100 != 0 || Year % 400 == 0

评估为

false && true && false || true

一步一步，然后计算为

((false && true) && false) || true
(false && false) || true
false || true
true

在该示例中，只要最后一个条件产生true，那么整个表达式也将产生true。我想这不是你想要的。

注意：我没有准确地花时间来完全验证您的逻辑。我只是发现了一个简单的错误，但可能还有其他错误。

Answer 2

您的代码中有大量重复。把它分开，让事情变得更容易。你的许多变量（一般来说，不仅仅是在编程术语中）只改变一次。只有当月份是二月时，年份才重要。

String MonthString;
int Days;
bool IsLeapYear = Year % 4 == 0 && (Year % 100 != 0 || Year % 400 == 0);
String Leap = "";
if (!IsLeapYear) {
    Leap = "not ";
}

if (MonthNumber == 1) {
    MonthString = "January";
    Days = 31;
}
else if (MonthNumber == 2) {
    MonthString = "February";
    if (IsLeapYear) {
        Days = 29;
    }
    else {
        Days = 28;
    }
}
else if (MonthNumber == 3) {
    MonthString = "March";
    Days = 31;
}
// etc....

JOptionPane.showMessageDialog(null, "The month is " + MonthString + ". " + 
    MonthString + " has " + Days + " days in it and it is " + Leap + 
    "a leap year.");

Answer 3

iFytil's answer 恰到好处。 这只是boolean algebra laws 的问题。我建议您阅读并了解基础知识。 这将有助于将来实现复杂条件，尤其是那些具有多个否定的条件。

此外，您的代码有些过于复杂。我不确定你的作业中是否有任何规则，但如果你分开处理这些事情会更干净。 与其同时评估两件事（哪个月和闰年），导致 12 x 2 条件，一个好的方法是将它分成两部分，导致 12 + 2 条件（如果在 2 月内有一点） .此外，当您有两个以上的比较时，switch-case 结构的可读性要高得多。

让我解释一下：

String answer;
switch (MonthNumber) {
    case 1:
        answer = "The month is January. January has 31 days in it.";
        break;
    case 2:
        answer = "The month is February. ";
        if (Year % 4 == 0 && !(Year % 100 == 0 && Year % 400 != 0))
            // years divisible by 4, except those divisible by 100 and not by 400
            answer += "February has 29 days in it because this is a leap-year.";
        else
            answer += "February has 28 days in it because this is not a leap-year.";
        break;
    case 3:
        ... // you get the idea
}

然后，紧随其后，解析闰年文本。

if (MonthNumber != 2) { // you have already completed the text in Feb
    if (Year % 4 == 0 && !(Year % 100 == 0 && Year % 400 != 0))
        answer += "and is also a leap-year.";
    else
        answer += "and it is not a leap-year.";
}

输出文本中的一个小变化将消除检查是否是二月的需要，但这只是一个技巧。

Answer 4

可以这样简化。

package leapmonth;

import javax.swing.JOptionPane;

public class LeapMonth {
    public static void main(String[] args) {

    String MonthInput;
    int YearInput;
    String[] MonthArray = {"January","February","March"
                            ,"April","May","June"
                            ,"July","August","September"
                            ,"October","November","December"
                          }; 
    // array always started from 0 then January should be 0

    JOptionPane.showMessageDialog(null, " HELLO, WELCOME TO LEAPMONTH ");
    Input = JOptionPane.showInputDialog( "Please enter the numerical respresentation of the month : ");
    MonthInput = Integer.parseInt(Input);
    Input = JOptionPane.showInputDialog( "Please enter the year : ");
    YearInput = Integer.parseInt(Input);

    if( MonthInput != 2 && Year % 4 == 0 && (Year % 100 != 0 || Year % 400 == 0))
    {
     JOptionPane.showMessageDialog(null,"The month is " + MonthArray[MonthInput+1] + ". " + MonthArray[MonthInput+1] + " has 31 days in it" + " and is also a leap year "); 
    }   
    else if( MonthInput != 2 && Year % 4 != 0 && (Year % 100 == 0 || Year % 400 != 0))
    {
     JOptionPane.showMessageDialog(null,"The month is " + MonthArray[MonthInput+1] + ". " + MonthArray[MonthInput+1] + " has 31 days in it" + " and it is not a leap year ");   
    }
    else if( MonthInput == 2 && Year % 4 == 0 && (Year % 100 != 0 || Year % 400 ==0))
    {
     JOptionPane.showMessageDialog(null,"The month is " + MonthArray[MonthInput+1] + ". " + MonthArray[MonthInput+1] + " has 29 days in it because this is a leap year ");  
    } 
    else if( MonthInput == 2 && Year % 4 != 0 && (Year % 100 == 0 || Year % 400 != 0))
    {
    JOptionPane.showMessageDialog(null,"The month is " + MonthArray[MonthInput+1] + ". " + MonthArray[MonthInput+1] + " has 28 days in it because this is not a leap year ");
    }
   }
}