2016-06-28 37 views
-2
import java.util.Scanner; 
public class hey { 



public static void main(String[] args) 
{ 
    Scanner in = new Scanner(System.in); 
    System.out.print("Please enter an integer between 6 and 12, inclusive: "); 
    int num = in.nextInt(); 
    System.out.print(num); 
    System.out.println(); 
    boolean result = shouldProcess(num); 
    processInput(result, num); // passing the result and num 

} 

public static boolean shouldProcess(int n) 
{ 
    if (n>=6 && n<12) 
    { 
     return true; 
    } 
    else 
    { 
     return false; 
    } 
} 

public static void processInput(boolean result2, int num) // added int num argument 
{ 
    if (result2 == true) 
    { 

      int sum = 0; 
      for (int i = 1; i <=num; i++) 
      { 
        sum +=i; 
      } 
      System.out.println("The sum from 1 to " + num+ " is: " +sum); 

    } 
    else 
    { 
     System.out.println("Number is outside of acceptable range"); 
    } 
} 
} 

这段代码在shouldProcess中返回一个布尔值,如果给定的数字介于6和12之间(包含),则返回true,如果不是则返回false。然后我创建了processInput方法,该方法使用shouldProcess来确定它是否可以计算和输出总和(只有当方法shouldProcess返回true时。找到总和的方法称为findSum。所以我不太确定如何进行下面的操作块,使这个findSum方法在我的代码:基本的java程序设计方法

public static void processInput(boolean result2, int num) 
{ 
    if (result2 == true) 
    { 

      int sum = 0; 
      for (int i = 1; i <=num; i++) 
      { 
        sum +=i; 
      } 
      System.out.println("The sum from 1 to " + num+ " is: " +sum); 

    } 
    else 
    { 
     System.out.println("Number is outside of acceptable range"); 
    } 
+0

但你已经建立了一个总和? – Fildor

+0

我该如何取得这一结果,并使其自己的方法称为findSum以下processInput – basicjava

+1

shouldProcess不包括在内。你说的少于12而不是<= 12 – Beto

回答

0
if (result2 == true) 
{ 

     int sum = 0; 
     for (int i = 1; i <=num; i++) 
     { 
       sum +=i; 
     } 
     System.out.println("The sum from 1 to " + num+ " is: " +sum); 

} 

将是:

if (result2 == true) 
{ 

     System.out.println("The sum from 1 to " + num+ " is: " +findSum(num)); 

} 

public static int findSum(num){ 
    int sum = 0; 
    for(int i = 1; i <= num; i++){ sum+= i; } // usually I'll use int i = 0 and add an offset in the loop. 
    return sum; 
} 

顺便说一句:你可以缩短你的过程:

public static boolean shouldProcess(int n) 
{ 
    return (n>=6 && n<=12); // both inclusive! 
} 
+0

我在哪里把公共静态诠释findSum(编号)编辑? – basicjava

+0

没关系没关系我非常感谢 – basicjava