Question

如何使大于100或小于0的数字不能平均添加？脚本编译并运行正常，但当我输入101时，它告诉我数字太大，但仍然取其平均值。谢谢！

import java.util.Scanner;

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

    Scanner dylan = new Scanner(System.in);

    int gradeCounter;
    int total;
    int grade;
    double average;

    total = 0;
    gradeCounter = 0;

    System.out.println( "Enter a grade or -1 to quit: " );
    grade = dylan.nextInt();

    while ( grade != -1 )
    {  


        total = total + grade;
        gradeCounter = gradeCounter + 1;




        if( grade > 100 || grade < 0) {         

        if ( grade > 100 ) {
        System.out.println( "Number too large, enter a new number");            
        }

        if ( grade < 0 ) {
        System.out.println( "Number too small, enter a new number");
        }

        }




        System.out.println( "Enter grade or -1 to quit: ");
        grade = dylan.nextInt();



    }


    if ( gradeCounter != 0 )

    {


        average = (double) total / gradeCounter;


        System.out.printf( "Class average is %.2f", average );



    }

    else
        System.out.println( "No grades were entered" );

}

}

Answer 1

添加continue以跳到下一个循环迭代。将计算移动到这些检查下方，以便在检查失败时不执行计算。

if ( grade > 100 ) {
    System.out.println( "Number too large, enter a new number");  
    continue;
}

if ( grade < 0 ) {
    System.out.println( "Number too small, enter a new number");
    continue;
}

total = total + grade;
gradeCounter = gradeCounter + 1;

Answer 2

您在测试值之前进行平均。将平均代码放在else语句中。

    if ( grade > 100 || grade < 0) {         
        if ( grade > 100 ) {
            System.out.println( "Number too large, enter a new number");            
        }
        if ( grade < 0 ) {
            System.out.println( "Number too small, enter a new number");
        }
    } else {
        total = total + grade;
        gradeCounter = gradeCounter + 1;
    }

嵌套控制结构等级平均计算器

2 个答案: