如果n& lt;'n'和'm'之间的奇数否则打印“打印不能完成” [英] Odd number between 'n' and 'm' if n< m otherwise print "printing can not be done "

问题描述

该计划的出口并未正确显示135.任何人都可以帮助我。输出应该在你没有输入但我的输出从1到M之间。



我尝试过：

THE OUT PUT OF THE PROGRAM IS NOT COMING CORRECT IT IS SHOWING 135. ANYONE CAN HELP ME OUT . THE OUTPUT SHOULD SHOULD BE BETWEEN THE NO YOU HAVE ENTERED BUT MY OUTPUT COMES FROM 1 TO M .

What I have tried:

import java.util.*;
public class amit7
{
    public static void main()
    {
        Scanner am=new Scanner(System.in);
        int n,m;
        System.out.println("Enter two no");
        n=am.nextInt();
        m=am.nextInt();
        if(n<m)
        {
        for(n;n<m;n++)
        {
            if (n%2!=0)
            
             System.out.print(n+"");
             
            } 
            System.out.println("");
        }
            else{ System.out.println("MESSAGE");
        }
    }

}

推荐答案

编译并不意味着你的代码是对！ ：笑：

将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。



所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。



首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:

Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16

然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。

所以，你可以查看代码和很明显，它在某处：

Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:

int Double(int value)
   {
   return value * value;
   }

一旦你知道可能出现的问题，就开始使用调试器找出原因。在你的行上设一个断点：

Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on your line:

for(n;n<m;n++)

并运行你的应用程序。在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。

如果没有，为什么不呢？它有何不同？



这是一项非常值得开发的技能，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改进！

and run your app. Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?

This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!

程序的'编译'版本

The 'compiling' version of your program

import java.util.*;
public class Foo
{
  public static void main(String args[])
  {
    Scanner am = new Scanner(System.in);
    int n,m;
    System.out.println("Enter two no");
    n = am.nextInt();
    m = am.nextInt();
    if( n < m )
    {
      for(; n < m; n++)
      {
        if ( n % 2 != 0)
          System.out.print( n + " ");
      }
      System.out.println("");
    }
    else
    {
      System.out.println("MESSAGE");
    }
  }
}

有效。



执行样本：

works.

Execution sample:

java Foo
Enter two no
5
11
5 7 9 

这篇关于如果n& lt;'n'和'm'之间的奇数否则打印“打印不能完成”的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！