Why are we using bufferedReader?

Question about the task

Under discussion

This is my code for the solution. I have always used scanner in the passed before I started taking this codeGym course. I understand that it probably wants me to use the bufferedReader, but from what I've seen the bufferedReader can only take a String value. Why do we want to use bufferedReader instead of Scanner? To me it makes more sense to use the nextInt(); followed by nextLine; rather than parsing the string to an int each time. I am also wondering why the main method has the

throws Exception

If anyone is able to explain this to me I would appreciate it, I am just trying to get a better understanding.

package club.pillowmc;

import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);

        String name = input.nextLine();
        int salary = input.nextInt();
        input.nextLine();
        int years = input.nextInt();
        input.nextLine();

        System.out.println(name + " will receive " + salary + " in " + years + " years.");
    }
}

Use the keyboard to separately enter the name, number1, and number2.
Display the following phrase:
<name> will receive <number1> in <number2> years.

Here's an example:
Nick will receive 10000 in 5 years.

Requirements:

The program should output text.
The program must read data from the keyboard.
The displayed text must contain the entered name.
The displayed text must contain the entered number1.
The displayed text must contain the entered number2.
The displayed text must fully match the task conditions.

package com.codegym.task.task03.task0319; /* Predictions */ import java.util.Scanner; public class Solution { public static void main(String[] args) throws Exception { Scanner input = new Scanner(System.in); String name = input.nextLine(); int salary = input.nextInt(); input.nextLine(); int years = input.nextInt(); input.nextLine(); System.out.println(name + " will receive " + salary + " in " + years + " years."); } }

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Agent Smith

Level 38

15 August 2020, 15:20

useful

1. Exceptions will be covered later, but in short, here is a simple explanation: An exception is an event that occurs during the execution of a program that disrupts the normal flow of instructions (). In Java, all exceptions are represented by classes. Thus, when an exception occurs in a program, an object of some type of exception class is generated. You will be able to catch such exceptions and handle them as you see fit. We have to include that piece of code there because otherwise the program won't compile, but to fully understand how it works you first need to gain more basic knowledge. So you can just ignore it for now or try reading Java - Exceptions 2. I think they want you to name your variables accordingly: name, number1, number2 to pass the test, that's what I did in my solution. But I'm not 100% sure, since I didn't try using custom names. 3. Why do you have input.NextLine() in lines 17 and 19? The program seems to work the same way without them. 4. Here is a nice little article about difference between Scanner and BufferReader Classes I've seen people using both classes here without problem, so I don't think the auto-grader cares about which one you use if it is not explicitly stated in the description.

Chase

Level 17

15 August 2020, 18:02

I used nextLine(); because in my other Java course I was told that when you get an integer value from the scanner it is always good practice just in case you need a string later in the program. I ended up just re coding the program the same way with the bufferedReader class with the same variable names, It seemed to work just fine. Still cant seem to figure out why I got an error on the task, because everything worked fine on IntelliJ.

Chase

Level 17

15 August 2020, 18:18

Here is an example of why I used the nextLine();

Scanner input = new Scanner(System.in);

int years = input.nextInt();
input.nextLine();
String name = input.nextLine();

System.out.println(name + " will take over the world in " + years + " years. Mwa-ha-ha!");

If you run this code without the nextLine(); you will notice that the program skips initializing the variable name and instead just prints a message without a name at all.

Chase

Level 17

15 August 2020, 14:47

I wanted to mention that bufferedReader does actually take an integer value, however it does still need to be parsed because when I enter a value such as 65, I get an output of 54. I am very confused about bufferedReader...

Nouser

Level 36 , Germany

15 August 2020, 16:56

useful

you probably used BufferedReaders read method. This reads exactly one byte, in your example the 6 from 65. And the ascii code of 6 is 54. If you convert the read byte to a char and output that you have again the 6. You should have a look at the method readLine... like

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
String input = reader.readLine();
int number1 = Integer.parseInt(input);
//or
int number2 = Integer.parseInt(reader.readLine());

Chase

Level 17

15 August 2020, 18:07

In my example I used an Integer value. I only have a few options. The reason why I wanted to use the Scanner class instead is so that I could use Int values instead of parsing strings to an int each time. Hopefully that made sense However that does make sense. Thanks for the information.

Nouser

Level 36 , Germany

16 August 2020, 10:20

What you want to use is up to you. - Scanner class is using it's own exception handling (it hides IOException so you do not need throws or try). This may not always be wanted. - BufferdReader just reads the stream up to the next linebreak or a single char while Scanner is parsing for tokens (or even regex). - BufferedReader is synchronized In the end it's up to you and depends on the code you want to write.