这是一个带有测试人员类的两部分代码。出于某种原因,我无法弄清楚如何将此字符串变量转换为正确参与等式的函数gradeNum = grad - .3和gradeNum = grade + 0.3。有什么想法吗?
/** This is my class Grade */
public class Grade
{
private String grade;
private double gradeNum;
// Constructor
public Grade(String showGrade)
{
grade = showGrade;
gradeNum = 0;
}
// getNumericGrade method to return grade number
public double getNumericGrade()
{
if (grade.equalsIgnoreCase("A") || grade.equalsIgnoreCase("A+"))
{
gradeNum = 4.0;
}
else if (grade.equalsIgnoreCase("A-"))
{
gradeNum = 3.7;
}
if (grade.equalsIgnoreCase("B"))
{
gradeNum = 3.0;
}
else if (grade.equalsIgnoreCase("B+"))
{
gradeNum = 3.3;
}
else if (grade.equalsIgnoreCase("B-"))
{
gradeNum = 2.7;
}
if (grade.equalsIgnoreCase("C"))
{
gradeNum = 2.0;
}
else if (grade.equalsIgnoreCase("C+"))
{
gradeNum = 2.3;
}
else if (grade.equalsIgnoreCase("C-"))
{
gradeNum = 1.7;
}
if (grade.equalsIgnoreCase("D"))
{
gradeNum = 1.0;
}
else if (grade.equalsIgnoreCase("D+"))
{
gradeNum = 1.3;
}
else if (grade.equalsIgnoreCase("D-"))
{
gradeNum = .7;
}
if (grade.equalsIgnoreCase("F"))
{
gradeNum = 0.0;
}
return gradeNum;
System.out.println("Invalid letter grade");
return -1.0;
}
}
/** This is my class tester
*/
import java. util .Scanner;
/**
This program tests the numeric grade with a given letter grade.
*/
public class GradeTester
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
System.out.println("Enter a letter grade: ");
String showGrade = in.nextLine();
Grade gg = new Grade(showGrade);
System.out.print("Numeric Value: ");
System.out.println(gg.getNumericGrade());
in.close();
}
}
答案 0 :(得分:1)
大卫的&肯的答案是正确的。您需要同时实现它们,然后另外删除一些return语句,然后代码将按预期工作。这是一个例子:
if (Character.toUpperCase(grade.charAt(0)) == 'B')
gradeNum = 3.0;
if (grade.length() > 1 && Character.toUpperCase(grade.charAt(1)) == '-')
gradeNum = gradeNum - 0.3;
return gradeNum;
之前,由于您正在使用grade.equalsIgnoreCase("B")
进行检查,因此输入的内容从未匹配为“B-”。因此,gradeNum
永远不会有3.0
的值。它仍为0.0
,然后执行gradeNum = gradeNum - 0.3;
,结果为-0.3
。
此外,您可以使用switch case而不是if语句。它会更优雅/可读性更强高效。
答案 1 :(得分:0)
使用:
而不是if (grade.equalsIgnoreCase("B"))
等
if (Character.toUpperCase(grade.charAt(0)) == 'B')
而不是grade.equalsIgnoreCase("A-")
等,请使用:
if (grade.length() > 1 && grade.charAt(1) == '-')
这足以让您解决问题。
注意:还需要进行其他更改,但我不想只是为您提供正确的代码。