使用void方法格式化java文本

Question

我已查找並找不到要查找的信息。我的代碼正如我所期望的那樣工作，但我有一點我想要改進的代碼。

的問題是，我不能像這樣的打印語句中調用一個void方法：

System.out.print("Water is a " + printTemp(temperature) + " at" + temperature + " degrees."; 

printTemp（溫度是一個void方法，以便這不會工作，結果我發現了一個工作-around但它並不理想：

System.out.print("\nWater is a "); 
printTemp(temperature); 
System.out.print(" at"); 
System.out.printf(" %.0f", temperature); 
System.out.print(" degrees.\n"); 

這裏是全碼：

import java.util.Scanner; 
public class printTemp { 

public static void main(String[]args) { 

Scanner input = new Scanner(System.in); 
System.out.print("Please enter the temperature: "); 
Double temperature = input.nextDouble(); 
// takes the state of the water from the printTemp method and 
// the temperature to return a formatted output to the user 
System.out.print("\nWater is a "); 
printTemp(temperature); 
System.out.print(" at"); 
System.out.printf(" %.0f", temperature); 
System.out.print(" degrees.\n"); 

} 

public static void printTemp(double temperature) { 
    String returnMessage = "null" ; 

    if (temperature < 32) 
     returnMessage = "solid"; 
    else if (temperature > 212) 
     returnMessage = "gas"; 
    else 
     returnMessage = "liquid"; 
    System.out.printf(returnMessage); 
} 

}

這是學校因此有條件必須保持，printTemp必須是一個無效的方法和變量溫度必須保持一個DOUBLE。

Answer 1

怎麼樣把下面的代碼片段

... 
System.out.print("Water is a " + returnMessage + " at" + temperature + " degrees."); 

到printTemp方法的最後一行？然後在main輸出什麼，你只是寫：

... 
printTemp(input.nextDouble()); 

Answer 2

使用類變量來保存溫度串詞，並使用無效的方法來設置它。

public class PrintTemp { 

    private static String TempStr = ""; 

    public static void main(String[]args) { 
     [...] 
     printTemp(temperature); 
     System.out.print("Water is a " + TempStr + " at" + temperature + " degrees."); 
    } 

    public static void printTemp(double temperature) { 
     if (temperature < 32) 
      TempStr = "solid"; 
     else if (temperature > 212) 
      TempStr = "gas"; 
     else 
      TempStr = "liquid"; 
    } 
} 

Answer 3

UPDATE

public class PrintTemp { 

    public static void main(String[] args) { 
     double temperature = 35; 
     StringBuilder returnMessage = new StringBuilder(); 
     printTemp(temperature, returnMessage); 
     System.out.print("Water is a " + returnMessage + " at " + temperature + " degrees."); 
    } 

    public static void printTemp(double temperature, StringBuilder returnMessage) { 
     if (temperature < 32) 
      returnMessage.append("solid"); 
     else if (temperature > 212) 
      returnMessage.append("gas"); 
     else 
      returnMessage.append("liquid"); 
    } 
} 

這是有printTemp修改消息沒有實際它返回，並且不使用額外的類/豆最簡單的方法。 但是，這隻適用於對象（不是原始類型），對象之間不適用於字符串，因爲在Java中它們被視爲一種特殊情況。 （這就是爲什麼我不得不使用一個StringBuilder，StringBuffer也可以）

Answer 4

對不起，推遲傢伙！

以下是解決方案。

import java.util.Scanner; 
public class printTemp { 
public static void main(String[]args) { 
//Read in the temperature from the user 
Scanner input = new Scanner(System.in); 
System.out.print("\nPlease enter the temperature: "); 
//Call and insert the input into the printTemp method. 
printTemp(input.nextDouble()); 
} 
public static void printTemp(double temperature) { 
    //Decide whether the water is in a soild, liquid or gaseous sate. 
    String returnMessage = "null"; 
    if (temperature < 32) 
     returnMessage = "solid"; 
    else if (temperature > 212) 
     returnMessage = "gas"; 
    else 
     returnMessage = "liquid"; 
    //Print to the user. 
    System.out.print("\nWater is a " + returnMessage); 
    System.out.printf(" at %.0f degrees.%n\n", temperature); 
} 

}