2016-06-28 66 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"); 
    } 
} 
} 

這段代碼在shouldProcess中返回一個布爾值,如果給定的數字介於6和12之間(包含),則返回true,如果不是則返回false。然後我創建了processInput方法,該方法使用shouldProcess來確定它是否可以計算和輸出總和(只有當方法shouldProcess返回true時。找到總和的方法稱爲findSum。所以我不太確定如何進行下面的操作塊,使這個findSum方法在我的代碼:基本的java程序設計方法

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

但你已經建立了一個總和? – Fildor

+0

我該如何取得這一結果,並使其自己的方法稱爲findSum以下processInput – basicjava

+1

shouldProcess不包括在內。你說的少於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); 

} 

將是:

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; 
} 

順便說一句:你可以縮短你的過程:

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

我在哪裏把公共靜態詮釋findSum(編號)編輯? – basicjava

+0

沒關係沒關係我非常感謝 – basicjava