Java将扫描器传递给方法 [英] Java passing scanner into method
问题描述
因此，我正在研究一些基本的编程思想，以帮助我掌握Java的知识，我创建了一个程序，该程序可以将PI打印到小数点后十位(如果需要，我可以添加更多内容). /p>
但是，我决定走一个额外的步骤，并创建一个选项来使程序不断运行，直到用户告诉它停止为止.我创建了一个返回true(再次运行)或false(退出程序)的方法.最初，我在该方法中创建了一个扫描仪以接受用户输入，并且程序以这种方式运行良好，但是它告诉我我没有在该方法中关闭扫描仪，导致资源泄漏.
我刚刚将输入扫描程序从main传递给方法作为参数，但是当我运行该程序时，它不接受用户输入，并且会打印出对不起，有一个错误"(else {}选项)在我的方法中的if-else语句中).现在，我可以回去创建一个单独的扫描仪，但是我的OCD不想Eclipse告诉我资源泄漏(我认为input.close()关闭了这两个扫描仪，但我不确定).
这是我的代码，对于那些对我不了解，正在学习的不良做法感到迷恋和冒犯的Java爱好者，我深表歉意.
import java.util.Scanner;
import java.text.DecimalFormat;
public class PiDecimalFormat {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
DecimalFormat format = new DecimalFormat("#");
int decPlace = 0;
boolean runProgram = true;
System.out.println("This program will print out PI to the decimal place of your choosing.");
while (runProgram == true) {
System.out.print("\nEnter the number of decimal places (up to 10) that \nyou would like to print PI to: ");
decPlace = input.nextInt();
switch (decPlace) {
case 0:
format = new DecimalFormat("#");
break;
case 1:
format = new DecimalFormat("#.#");
break;
case 2:
format = new DecimalFormat("#.##");
break;
case 3:
format = new DecimalFormat("#.###");
break;
case 4:
format = new DecimalFormat("#.####");
break;
case 5:
format = new DecimalFormat("#.#####");
break;
case 6:
format = new DecimalFormat("#.######");
break;
case 7:
format = new DecimalFormat("#.#######");
break;
case 8:
format = new DecimalFormat("#.########");
break;
case 9:
format = new DecimalFormat("#.#########");
break;
case 10:
format = new DecimalFormat("#.##########");
break;
}
System.out.println("\nThe value of PI to " + decPlace + " decimal places is " + format.format(Math.PI) + ".");
runProgram = AskRunAgain(input);
}
input.close();
}
static boolean AskRunAgain(Scanner askUser) {
String userChoice;
System.out.print("\nWould you like to run the program again? [y/n]: ");
userChoice = askUser.nextLine();
if ((userChoice.equals("y")) || (userChoice.equals("Y")) || (userChoice.equals("yes")) ||
(userChoice.equals("Yes")) || (userChoice.equals("YES"))) {
return true;
}
else if ((userChoice.equals("n")) || (userChoice.equals("N")) || (userChoice.equals("no")) ||
(userChoice.equals("No")) || (userChoice.equals("NO"))) {
System.out.println("\nExitting the program. have a good day!");
return false;
}
else {
System.out.println("Sorry, there was an error.");
return false;
}
}
}
如果有人能告诉我为什么这样做，我将不胜感激.我是Java的新手(C/C ++/C#和Python不错).我没有看到有关此特定问题的其他问题，如果我仅在该方法中创建另一个扫描仪，那也没什么大不了的.
我注意到您正在执行此呼叫:
decPlace = input.nextInt();
不使用返回字符，因此就Scanner而言，它仍在缓冲区中.
这意味着，对于输入2\n，它将读取2作为下一个整数，但是读取一个空字符串以调用nextLine().
要克服这一点，请在读取下一个整数后使用input.nextLine()完成对行的使用.
decPlace = input.nextInt();
input.nextLine();
So I'm going through a bunch of basic programming ideas to help me get the hang of Java and I created a program that will print PI up to the 10th decimal place (and I can add more if I want).
However, I decided to go an extra step and create an option to keep running the program over and over until the user tells it to stop. I created a method to return either true (run again) or false (exit program). Originally I created a scanner in the method to take user input, and the program runs fine that way, but it tells me I have a resource leak because I didn't close the scanner in the method.
I have just passed the input scanner from main to the method as a parameter, but when I run the program it doesn't accept user input and will print out "Sorry, there was an error" (the else{} option in the if-else statement in my method). Now, I can go back and create a seperate scanner but my OCD doesn't want Eclipse telling me there is a resource leak (I think input.close() closes both scanners, but I am not sure).
Here is my code, I apologize to any Java enthusiasts who become smitten and offended at any bad practices I am unaware of, I am learning.
import java.util.Scanner;
import java.text.DecimalFormat;
public class PiDecimalFormat {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
DecimalFormat format = new DecimalFormat("#");
int decPlace = 0;
boolean runProgram = true;
System.out.println("This program will print out PI to the decimal place of your choosing.");
while (runProgram == true) {
System.out.print("\nEnter the number of decimal places (up to 10) that \nyou would like to print PI to: ");
decPlace = input.nextInt();
switch (decPlace) {
case 0:
format = new DecimalFormat("#");
break;
case 1:
format = new DecimalFormat("#.#");
break;
case 2:
format = new DecimalFormat("#.##");
break;
case 3:
format = new DecimalFormat("#.###");
break;
case 4:
format = new DecimalFormat("#.####");
break;
case 5:
format = new DecimalFormat("#.#####");
break;
case 6:
format = new DecimalFormat("#.######");
break;
case 7:
format = new DecimalFormat("#.#######");
break;
case 8:
format = new DecimalFormat("#.########");
break;
case 9:
format = new DecimalFormat("#.#########");
break;
case 10:
format = new DecimalFormat("#.##########");
break;
}
System.out.println("\nThe value of PI to " + decPlace + " decimal places is " + format.format(Math.PI) + ".");
runProgram = AskRunAgain(input);
}
input.close();
}
static boolean AskRunAgain(Scanner askUser) {
String userChoice;
System.out.print("\nWould you like to run the program again? [y/n]: ");
userChoice = askUser.nextLine();
if ((userChoice.equals("y")) || (userChoice.equals("Y")) || (userChoice.equals("yes")) ||
(userChoice.equals("Yes")) || (userChoice.equals("YES"))) {
return true;
}
else if ((userChoice.equals("n")) || (userChoice.equals("N")) || (userChoice.equals("no")) ||
(userChoice.equals("No")) || (userChoice.equals("NO"))) {
System.out.println("\nExitting the program. have a good day!");
return false;
}
else {
System.out.println("Sorry, there was an error.");
return false;
}
}
}
If anyone could tell me why it is doing this, I would appreciate it. I'm new to Java (decent with C/C++/C# and Python). I didn't see other questions about this specific problem, and it's not a big deal if I just create another scanner in the method.
I notice that you're doing this call:
decPlace = input.nextInt();
The return character isn't consumed, so it is still on the buffer as far as the Scanner is concerned.
This means, for an input of 2\n, it will read the 2 as the next integer, but read an empty string for the call to nextLine().
To get past this, finish consuming the line by using input.nextLine() after you read the next integer.
decPlace = input.nextInt();
input.nextLine();
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