我一直无法理解Java方法的概念，我需要编写一个等级计算器程序的代码，它比我的程序更简洁。
这就是我所拥有的（这是一种方法）：
public class GradeCalculator2{
public static void main(String[]args){
Scanner sc = new Scanner(System.in);
int students;
String studentName;
System.out.println("Welcome to GradeCalculator program.");
System.out.println("Author: Hayden Wires");
System.out.println("Date: 10/25/2016");
System.out.print("Enter the number of students:");
students = sc.nextInt();
System.out.println("Enter next student's information:");
System.out.print("Name:");
studentName = sc.next();
for(int i = students; i > 1; i++){
double quizScore1;
double quizScore2;
double quizScore3;
double quizScore4;
double quizScore5;
System.out.print("Quiz 1:");
quizScore1 = sc.nextDouble();
quizScore1 /= 100;
quizScore1 *= .02;
System.out.print("Quiz 2:");
quizScore2 = sc.nextDouble();
quizScore2 /= 100;
quizScore2 *= .02;
System.out.print("Quiz 3:");
quizScore3 = sc.nextDouble();
quizScore3 /= 100;
quizScore3 *= .02;
System.out.print("Quiz 4:");
quizScore4 = sc.nextDouble();
quizScore4 /= 100;
quizScore4 *= .02;
System.out.print("Quiz 5:");
quizScore5 = sc.nextDouble();
quizScore5 /= 100;
quizScore5 *= .02;
double totalQuizScores = quizScore1 + quizScore2 + quizScore3 + quizScore4 + quizScore5;
totalQuizScores /= 5;
double a1_Score;
double a2_Score;
double a3_Score;
double a4_Score;
double a5_Score;
double a6_Score;
System.out.print("Assignment 1:");
a1_Score = sc.nextDouble();
a1_Score /= 100;
a1_Score *= .025;
System.out.print("Assignment 2:");
a2_Score = sc.nextDouble();
a2_Score /= 100;
a2_Score *= .025;
System.out.print("Assignment 3:");
a3_Score = sc.nextDouble();
a3_Score /= 100;
a3_Score *= .05;
System.out.print("Assignment 4:");
a4_Score = sc.nextDouble();
a4_Score /= 100;
a4_Score *= .05;
System.out.print("Assignment 5:");
a5_Score = sc.nextDouble();
a5_Score /= 100;
a5_Score *= .05;
System.out.print("Assignment 6:");
a6_Score = sc.nextDouble();
a6_Score /= 100;
a6_Score *= .1;
double a1_a2_Scores = a1_Score + a2_Score;
double a3_a4_a5_Scores = a3_Score + a4_Score + a5_Score;
double assignmentScoresTotal = a1_a2_Scores + a3_a4_a5_Scores + a6_Score;
assignmentScoresTotal /= 3;
double mt1Score;
System.out.print("Midterm 1:");
mt1Score = sc.nextDouble();
mt1Score /= 100;
mt1Score *= .1;
double mt2Score;
System.out.print("Midterm 2:");
mt2Score = sc.nextDouble();
mt2Score /= 100;
mt2Score *= .2;
double mtScoresTotal = mt1Score + mt2Score;
mtScoresTotal /= 2;
double finalScore;
System.out.print("Final:");
finalScore = sc.nextDouble();
finalScore /= 100;
finalScore *= .3;
double totalGrade = assignmentScoresTotal + totalQuizScores + mtScoresTotal + finalScore;
totalGrade *= 100;
totalGrade *= 2;
System.out.println(studentName + "'S SCORE:" + totalGrade);
String letterGrade;
if (totalGrade >= 93){
letterGrade = "A";
} else if ((totalGrade >= 90)&&(totalGrade < 92)){
letterGrade = "A-";
} else if ((totalGrade >= 87)&&(totalGrade < 90)){
letterGrade = "B+";
} else if ((totalGrade >= 83)&&(totalGrade < 87)){
letterGrade = "B";
} else if ((totalGrade >= 80)&&(totalGrade < 83)){
letterGrade = "B-";
} else if ((totalGrade >= 77)&&(totalGrade < 80)){
letterGrade = "C+";
} else if ((totalGrade >= 74)&&(totalGrade < 77)){
letterGrade = "C";
} else if ((totalGrade >= 70)&&(totalGrade < 74)){
letterGrade = "C-";
} else if ((totalGrade >= 68)&&(totalGrade < 70)){
letterGrade = "D+";
} else if ((totalGrade >= 60)&&(totalGrade < 68)){
letterGrade = "D";
} else {
letterGrade = "F";
}
System.out.println(studentName + "'S LETTER GRADE:" + letterGrade);
}
}
}
我知道这非常混乱，可能很难阅读，但这就是我需要帮助的地方。我知道有一种方法可以使用方法减少冗余，我只是无法在这个程序的上下文中掌握这个概念。
答案 0 :(得分：2)
Java是精心设计的面向对象编程语言。面向对象的哲学与继承高度相关。 Java是在OOP的继承属性上构建的。由于继承，你不能拥有全局方法。
以下示例不是良好的面向对象实践，只是为了显示过程：
public class myClass{
public static void main(String[] args){
f1();
}
private void f1(arg1, arg2){
//your code here
} //end of method f1
} //end of class
这不是一个好的面向对象实践，因为你通常将你的主（驱动）类和其他类分开，而在主类中你只有主要的方法。
答案 1 :(得分：1)
要缩短代码，您需要寻找模式！
了解如何以类似的方式处理每个测验和每个作业的得分？
您总是按照以下方式处理测验：
System.out.print("Quiz <insert a number here>:");
quizScoreX = sc.nextDouble();
quizScoreX /= 100;
quizScoreX *= .02;
这里唯一不同的是“测验”一词之后的数字。
您总是按照以下方式处理作业：
System.out.print("Assignment <insert number here>:");
aX_Score = sc.nextDouble();
aX_Score /= 100;
aX_Score *= <insert number here>;
这里唯一不同的是“作业”一词之后的数字以及你将得分乘以的数字。
看模式？
这意味着我们可以将每个测验和每个作业的得分分为两种方法：
private static double getQuizScore(int number) {
System.out.print("Quiz " + Integer.toString(number) + ":");
double quizScore = sc.nextDouble();
quizScore /= 100;
quizScore *= .02;
return quizScore;
}
private static double getAssignmentScore(int number, double multiplier) {
System.out.print("Assignment " + Integer.toString(number) + ":");
doubke assignmentScore = sc.nextDouble();
assignmentScore /= 100;
assignmentScore *= multiplier;
return assignmentScore;
}
另请注意，您将获得5个测验的分数。每次，你正在做的只是递增测验号码。然后你将它们加在一起。这可以简化为：
private static double getTotalQuizScore() {
double total = 0;
for (int i = 0 ; i < 5 ; i++) {
total += getQuizScore(i + 1);
}
return total;
}
然后，您可以在代码中使用这两种方法：
double totalQuizScores = getTotalQuizScore();
totalQuizScores /= 5;
double a1_Score = getAssignmentScore(1, 0.25);
double a2_Score = getAssignmentScore(2, 0.25);
double a3_Score = getAssignmentScore(3, 0.05);
double a4_Score = getAssignmentScore(4, 0.05);
double a5_Score = getAssignmentScore(5, 0.05);
double a6_Score = getAssignmentScore(6, 0.1);
double a1_a2_Scores = a1_Score + a2_Score;
double a3_a4_a5_Scores = a3_Score + a4_Score + a5_Score;
double assignmentScoresTotal = a1_a2_Scores + a3_a4_a5_Scores + a6_Score;
assignmentScoresTotal /= 3;
double mt1Score;
System.out.print("Midterm 1:");
mt1Score = sc.nextDouble();
mt1Score /= 100;
mt1Score *= .1;
double mt2Score;
System.out.print("Midterm 2:");
mt2Score = sc.nextDouble();
mt2Score /= 100;
mt2Score *= .2;
double mtScoresTotal = mt1Score + mt2Score;
mtScoresTotal /= 2;
double finalScore;
System.out.print("Final:");
finalScore = sc.nextDouble();
finalScore /= 100;
finalScore *= .3;
double totalGrade = assignmentScoresTotal + totalQuizScores + mtScoresTotal + finalScore;
totalGrade *= 100;
totalGrade *= 2;
System.out.println(studentName + "'S SCORE:" + totalGrade);
String letterGrade;
if (totalGrade >= 93){
letterGrade = "A";
} else if ((totalGrade >= 90)&&(totalGrade < 92)){
letterGrade = "A-";
} else if ((totalGrade >= 87)&&(totalGrade < 90)){
letterGrade = "B+";
} else if ((totalGrade >= 83)&&(totalGrade < 87)){
letterGrade = "B";
} else if ((totalGrade >= 80)&&(totalGrade < 83)){
letterGrade = "B-";
} else if ((totalGrade >= 77)&&(totalGrade < 80)){
letterGrade = "C+";
} else if ((totalGrade >= 74)&&(totalGrade < 77)){
letterGrade = "C";
} else if ((totalGrade >= 70)&&(totalGrade < 74)){
letterGrade = "C-";
} else if ((totalGrade >= 68)&&(totalGrade < 70)){
letterGrade = "D+";
} else if ((totalGrade >= 60)&&(totalGrade < 68)){
letterGrade = "D";
} else {
letterGrade = "F";
}
System.out.println(studentName + "'S LETTER GRADE:" + letterGrade);
编辑：我忘了提及，你还需要将sc移到类级别，以便方法可以访问它：
移动此行：
Scanner sc = new Scanner(System.in);
在主方法之外，并在开头添加private static：
private static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
// ...
}
答案 2 :(得分：1)
您似乎重复运行相同的计算：
System.out.print("Quiz 1:");
quizScore1 = sc.nextDouble();
quizScore1 /= 100;
quizScore1 *= .02;
这可以用方法
代替double readScore(Scanner sc, String title, double weight)
{
System.out.print(title);
double result = sc.nextDouble();
result /= 100.0;
result *= weight;
return result;
}
然后，你的主要功能如下：
quizScore1 = readScore(sc, "Quiz 1", 0.02);
:
a2_Score = readScore(sc, "Assignment 2", 0.025);
etc...
这样就可以从主函数中删除大量重复的代码。
之后，你应该学习数组，所以你没有a1_score到a6_score等...