在java上按一定量递减_Java_Loops_Subtraction_Decrement

在java上按一定量递减

java loops

在java上按一定量递减,java,loops,subtraction,decrement,Java,Loops,Subtraction,Decrement,我试图创建一个程序，用户输入的数字会减少一定数量。这是我想要的： Update by (Increment/Decrement):decrement Enter starting number:15 Enter update number:3 Enter end number:3 loop#1 value=15 loop#2 value=12 public class loop { public enum Operation {INCREMENT, DECREMENT, INVA

我试图创建一个程序，用户输入的数字会减少一定数量。这是我想要的：

Update by (Increment/Decrement):decrement
Enter starting number:15
Enter update number:3
Enter end number:3
loop#1  value=15
loop#2  value=12

public class loop {

    public enum Operation {INCREMENT, DECREMENT, INVALID}

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        System.out.print("Update by (Increment/Decrement):");
        String operationString = scan.nextLine();

        System.out.print("Enter starting number:");
        String numberString = scan.nextLine();

        System.out.print("Enter update number:");
        String upnumberString = scan.nextLine();

        System.out.print("Enter end number:");
        String endnumberString = scan.nextLine();

        // Determine and parse stuff
        int startNumber = Integer.parseInt(numberString);
        int updateNumber = Integer.parseInt(upnumberString);
        int endNumber = Integer.parseInt(endnumberString);
        // Parse operation, but assume invalid operation
        Operation operation = Operation.INVALID;
        if (operationString.equalsIgnoreCase("increment")) {
            operation = Operation.INCREMENT;
        } else if (operationString.equalsIgnoreCase("decrement")) {
            operation = Operation.DECREMENT;
        }

        // now do the "meat" of the assignment
        int loopNumber = 0; // we'll keep the loop number as a separate counter
        switch (operation) {
            case INCREMENT:
                for (int i = startNumber; i < endNumber; i = i + updateNumber) {
                    loopNumber++;
                    performAssignmentPrinting(loopNumber, i);
                }
                break;

            case DECREMENT:
                for (int i = startNumber; i > endNumber; i = i - updateNumber) {
                    loopNumber++;
                    performAssignmentPrinting(loopNumber, i)
                }
                break;

            case INVALID:
            default:
                throw new IllegalStateException("Please enter supported operation! (increment/decrement)");
        }
    }

    private static void performAssignmentPrinting(int loopNumber, int value) {
        System.out.println("loop#" + loopNumber + "\tvalue=" + value);
    }
}

结束编号是循环将停止的位置，更新编号是起始编号应减少多少。到目前为止，这是我的代码，我被困在如何保持循环，直到结束数字

package jaba;
import java.util.Scanner;


public class loop {

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    Scanner input = new Scanner(System.in);

    System.out.print("Update by (Increment/Decrement):");
    String Decrement = scan.nextLine();
    
    System.out.print("Enter starting number:");
    String number = scan.nextLine();
    
    System.out.print("Enter update number:");
    String upnumber = scan.nextLine();
    
    System.out.print("Enter end number:");
    String endnumber = scan.nextLine();
    
    
    
    int i,j;
    
    i = 15;
    j = 1;
    
    do {
        
        System.out.println("loop#" +j+ "\tvalue="+i);
        j++;
        
        
        
        }while(i<15);
            
        i = i-3;
        
        System.out.println("loop#" +j+ "\tvalue="+i);
        };
        
        
}

包jaba；
导入java.util.Scanner；
公共类循环{
公共静态void main（字符串[]args）{
扫描仪扫描=新扫描仪（System.in）；
扫描仪输入=新扫描仪（System.in）；
系统输出打印（“更新方式（增量/减量）：”；
字符串减量=scan.nextLine（）；
系统输出打印（“输入起始编号：”）；
字符串编号=scan.nextLine（）；
系统输出打印（“输入更新编号：”）；
字符串upnumber=scan.nextLine（）；
系统输出打印（“输入结束编号：”）；
String endnumber=scan.nextLine（）；
int i，j；
i=15；
j=1；
做{
System.out.println（“循环#“+j+”\tvalue=“+i”）；
j++；
}而（iFor循环是程序的解决方案。For（a；b；c）{…}


谷歌搜索for循环是如何工作的。试着了解a、b、c三部分是如何工作的
伪：
// if decrease mode
    // for (i = upperbound ; i >= lowerbound ; i-= decrement)
        // print i

// if increase mode
    // for (i = lowerbound ; i <= upperbound ; i+= increment)
        // print i

//如果减少模式
//for（i=上限；i>=下限；i-=减量）
//打印i
//如果增加模式
//for（i=lowerbound；ifor循环是程序的解决方案。for（a；b；c）{…}


谷歌搜索for循环是如何工作的。试着了解a、b、c三部分是如何工作的
伪：
// if decrease mode
    // for (i = upperbound ; i >= lowerbound ; i-= decrement)
        // print i

// if increase mode
    // for (i = lowerbound ; i <= upperbound ; i+= increment)
        // print i

//如果减少模式
//for（i=上限；i>=下限；i-=减量）
//打印i
//如果增加模式
//对于（i=lowerbound；i这样的东西怎么样：
Update by (Increment/Decrement):decrement
Enter starting number:15
Enter update number:3
Enter end number:3
loop#1  value=15
loop#2  value=12

public class loop {

    public enum Operation {INCREMENT, DECREMENT, INVALID}

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        System.out.print("Update by (Increment/Decrement):");
        String operationString = scan.nextLine();

        System.out.print("Enter starting number:");
        String numberString = scan.nextLine();

        System.out.print("Enter update number:");
        String upnumberString = scan.nextLine();

        System.out.print("Enter end number:");
        String endnumberString = scan.nextLine();

        // Determine and parse stuff
        int startNumber = Integer.parseInt(numberString);
        int updateNumber = Integer.parseInt(upnumberString);
        int endNumber = Integer.parseInt(endnumberString);
        // Parse operation, but assume invalid operation
        Operation operation = Operation.INVALID;
        if (operationString.equalsIgnoreCase("increment")) {
            operation = Operation.INCREMENT;
        } else if (operationString.equalsIgnoreCase("decrement")) {
            operation = Operation.DECREMENT;
        }

        // now do the "meat" of the assignment
        int loopNumber = 0; // we'll keep the loop number as a separate counter
        switch (operation) {
            case INCREMENT:
                for (int i = startNumber; i < endNumber; i = i + updateNumber) {
                    loopNumber++;
                    performAssignmentPrinting(loopNumber, i);
                }
                break;

            case DECREMENT:
                for (int i = startNumber; i > endNumber; i = i - updateNumber) {
                    loopNumber++;
                    performAssignmentPrinting(loopNumber, i)
                }
                break;

            case INVALID:
            default:
                throw new IllegalStateException("Please enter supported operation! (increment/decrement)");
        }
    }

    private static void performAssignmentPrinting(int loopNumber, int value) {
        System.out.println("loop#" + loopNumber + "\tvalue=" + value);
    }
}

像这样的怎么样：
Update by (Increment/Decrement):decrement
Enter starting number:15
Enter update number:3
Enter end number:3
loop#1  value=15
loop#2  value=12

public class loop {

    public enum Operation {INCREMENT, DECREMENT, INVALID}

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        System.out.print("Update by (Increment/Decrement):");
        String operationString = scan.nextLine();

        System.out.print("Enter starting number:");
        String numberString = scan.nextLine();

        System.out.print("Enter update number:");
        String upnumberString = scan.nextLine();

        System.out.print("Enter end number:");
        String endnumberString = scan.nextLine();

        // Determine and parse stuff
        int startNumber = Integer.parseInt(numberString);
        int updateNumber = Integer.parseInt(upnumberString);
        int endNumber = Integer.parseInt(endnumberString);
        // Parse operation, but assume invalid operation
        Operation operation = Operation.INVALID;
        if (operationString.equalsIgnoreCase("increment")) {
            operation = Operation.INCREMENT;
        } else if (operationString.equalsIgnoreCase("decrement")) {
            operation = Operation.DECREMENT;
        }

        // now do the "meat" of the assignment
        int loopNumber = 0; // we'll keep the loop number as a separate counter
        switch (operation) {
            case INCREMENT:
                for (int i = startNumber; i < endNumber; i = i + updateNumber) {
                    loopNumber++;
                    performAssignmentPrinting(loopNumber, i);
                }
                break;

            case DECREMENT:
                for (int i = startNumber; i > endNumber; i = i - updateNumber) {
                    loopNumber++;
                    performAssignmentPrinting(loopNumber, i)
                }
                break;

            case INVALID:
            default:
                throw new IllegalStateException("Please enter supported operation! (increment/decrement)");
        }
    }

    private static void performAssignmentPrinting(int loopNumber, int value) {
        System.out.println("loop#" + loopNumber + "\tvalue=" + value);
    }
}

您有i=i-3；out-of-loop。
将i的递减移动到循环中：
    do {
        System.out.println("loop#" + j + "\tvalue=" + i);
        j++;
        i = i - 3;
    } while (i > endnumber);

您有i=i-3；out-of-loop。
将i的递减移动到循环中：
    do {
        System.out.println("loop#" + j + "\tvalue=" + i);
        j++;
        i = i - 3;
    } while (i > endnumber);

提示：您的do/while
循环有一个取决于i
的终止条件，但循环体不会更改i
…因此它将只执行一次或永远执行。提示：您的do/while
循环有一个取决于i
的终止条件，但循环体不会更改E>我<或代码>…所以它将执行一次或全部。当我理解这是一个学校作业时，考虑下面的输入：<代码> StastNox= 1000 ，<代码> EndoNo.= -1000 < <代码>，<代码> UpDeNoNoST＝100 < /Cord>，并查看当您输入增量和减量时会发生什么。注意循环数WH的差异。如果我没弄错的话，它应该在21474846左右。CC@coderx这个解决方案可能对我上面提到的输入也不起作用，因为它假定在递增的情况下，开始数将小于结束数，反之亦然，递减。@coderx添加了do/while版本，也应该对c起作用请看我在第一条评论中提到的示例。您好！非常感谢您提供的附加信息和源代码：）源代码确实帮助我理解了作业的逻辑和概念！再次感谢您！@ CordEX为加分，添加一个输入验证检查来警告用户，如果他输入“代码＞0”/代码>更新编号，并提示输入一次。我理解这是一个学校作业，考虑以下输入：<代码>>startNumber=1000
，endNumber=-1000
，updateNumber=100
，然后查看当您输入递增和递减时会发生什么。请注意递增时循环数的差异。如果我没有弄错的话，它应该在21474846左右。CC@CoderX此解决方案也可能不适用于我所述的输入above，因为它假定在递增的情况下，开始数将小于结束数，反之亦然，递减。@coderx添加了do/while版本，该版本也适用于我在第一条评论中所述的反例。您好！非常感谢您提供的附加信息和源代码：）源代码真的帮助我理解了分配的逻辑和概念！再次非常感谢！@coderx获得额外分数，添加输入验证检查，如果用户输入0
以获取更新编号，则警告用户，并再次提示输入。