CodeGym/Help with Java Tasks/Help please

Level 22

01.09.2020
810views
17comments

Help please

Question about the task

Resolved

Seems to work until a String is entered... But fails all tests...

1. Read data from the console until the word "exit" is entered.
2. For each value (except "exit"), call the print method. If the value:
2.1. contains a period ("."), then call the print method for Double;
2.2. is greater than zero, but less than 128, then call the print method for short;
2.3. less than or equal to zero or greater than or equal to 128, then call the print method for Integer;
2.4. otherwise, call the print method for String.

Requirements:

The program must read data from the keyboard.
The program should stop reading data from the keyboard after "exit" is entered.
If the entered string contains a period (".") and can be correctly converted to a Double, the print(Double value) method should be called.
If the entered string can be correctly converted to a short and the resulting number is greater than 0, but less than 128, the print(short value) method should be called.
If the entered string can be correctly converted to an Integer and the resulting number is less than or equal to 0 or greater than or equal to 128, the print(Integer value) method should be called.
In all other cases, the print(String value) method should be called.

package com.codegym.task.task15.task1519; import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; /* Different methods for different types */ public class Solution { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); while (true) { String input = reader.readLine(); if (input.equals("exit")) { break; } else if (input.contains(".")) { print(Double.parseDouble(input)); continue; } if (Integer.parseInt(input) > 0 && Integer.parseInt(input) < 128) { print((short) Integer.parseInt(input)); continue; } if (Integer.parseInt(input) <= 0 || Integer.parseInt(input) >= 128) { print(Integer.parseInt(input)); continue; } else { print(input); continue; } } } public static void print(Double value) { System.out.println("This is a Double. Value: " + value); } public static void print(String value) { System.out.println("This is a String. Value: " + value); } public static void print(short value) { System.out.println("This is a short. Value: " + value); } public static void print(Integer value) { System.out.println("This is an Integer. Value: " + value); } }

Comments (17)

Popular
New
Old

You must be signed in to leave a comment

Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2020, 13:22

solution

The problem is that Integer.parseInt() will throw a NumberFormatException whenever an invalid entry is made, such as the user entering anything other than a number (and "exit" because that will be caught at line 18). This is causing your program to error out on line 25 and the validation can not check any further. Everything else is working as intended. To complete this task you are going to need to figure out how to prevent this exception from either occurring in the first place, or handle it properly after it has occurred. *hint* if an exception occurs in a try/catch block, the code will move to the catch block and execute what is there. As mentioned above, there is only one way an exception would occur. Also, the Character class has a method called isDigit() that takes a char and will return true if it is a number and false otherwise. One of these techniques could solve the problem you are having. *Tips* In your code you have Integer.parseInt(input) 4 different times in the same block of code. This means, logically, the computer has the potential of using its resources to do the same exact operation 3 times. Imagine organizing your office, then dumping everything out and re-organizing the same exact way, then doing that again. It makes no sense. A better way of doing it would be to only do the operation once and storing that value to a variable. This is called 'caching'. So: Instead of:

if(Integer.parseInt(input) /*some operation*/ || Integer.parseInt(input) /*some operation*/)
if(Integer.parseInt(input) /*some operation*/ && Integer.parseInt(input) /*some operation*/)

do this:

int result = Integer.parseInt(input); // caching the value
if(result /*some operation*/ && result /*some operation*/)
if(result /*some operation*/ || result /*some operation*/)

Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2020, 13:47

Further, the two if statements that are responsible for separating ints and shorts (line 25 and line 29) are polar opposites of each other. That means if one is true the other is false, and vice versa. This means that they can be completely replaced by an if-else statement:

if(result /*some operation*/ && result /*some operation*/)
   // code if true
else
   // code if false

Now that you are getting higher in level you should be coding like this.

Seferi

Level 22 , United Federation of Planets

1 September 2020, 19:25

Thanks mate. Great tips!

Seferi

Level 22 , United Federation of Planets

1 September 2020, 19:57

Still can't get passed tho...

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

        while (true) {
            String input = reader.readLine();

            try {
                int intInput = Integer.parseInt(input);
                if (input.contains(".")) {
                    print(Double.parseDouble(input));
                    continue;
                }
                if (intInput > 0 && intInput < 128) {
                    print((short) intInput);
                } else {
                    print(intInput);
                }
            } catch (NumberFormatException e) {
                if (input.equals("exit")) {
                    break; }
                    else {
                        print(input);
                    }
                }

            }
        }

    public static void print(Double value) {
        System.out.println("This is a Double. Value: " + value);
    }

    public static void print(String value) {
        System.out.println("This is a String. Value: " + value);
    }

    public static void print(short value) {
        System.out.println("This is a short. Value: " + value);
    }

    public static void print(Integer value) {
        System.out.println("This is an Integer. Value: " + value);
    }
}

Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2020, 20:20

Great job cleaning up the code and making it more efficient on the logic. The problem now is line 9. If you enter a string that has a decimal, meaning its a double, then this line will throw a NumberFormatException as Integers can not have decimals in them. Consider moving this line to a location that would eliminate the possibility of this occurring.

Seferi

Level 22 , United Federation of Planets

2 September 2020, 19:16

Thanks mate! Finally got it to pass.. It was so obvious when you pointed out.. I should stop hurrying, take my time and better analize/debug the code..

Guadalupe Gagnon

Level 37 , Tampa, United States

2 September 2020, 19:27

Learning a computer language is really easy. Learning how to actually speak to a computer in logic is not. When you first learn how to program, 85% of that effort is just learning this computer logic and how to break down problems in very easy step by step instruction. I personally struggled myself missing a step here or there on the instructions to a complete solution. It's not easy to visualize how data can be logically modified inside a loop that loops a million times in 3/100's of a second. You are doing a good job though and learning quite quickly. Next year you will be looking back at these beginner examples laughing how easy they are when you have experience.

The Chief

Level 18 , United States

7 July 2021, 19:32

G, where should line 9 be moved?

Guadalupe Gagnon

Level 37 , Tampa, United States

7 July 2021, 20:11

Integer.parseInt() will throw an exception when anything that is not a valid integer character (such as letters, punctuation, decimal points, etc) are contained within the String being parsed. Logically, because decimal points can result in an exception being thrown when parsing to an int and this task says that the input can contain fractional numbers, you should first check if the number is fractional before parsing to a number type that can't hold fractional values.

dhanush

Level 30 , Chennai, India

1 September 2020, 12:01

useful

1# Use try catch because when you entered a String it will throw a numberformatexception it will be a problem,print(input) should be in the catch block. 2# All the if statement should be if...else statement. 3# remove all continue keyword (Still not verify the validation).

Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2020, 13:24

The use of the 'continue' key word in the code above works exactly as if they were if-else statements. If a line is reached with the 'continue' keyword, the loop (line 16) will end its current cycle and start a new one. This is a valid use of the structure.

dhanush

Level 30 , Chennai, India

1 September 2020, 14:04

Thanks for that,i exactly don't know what the purpose of continue like break and return statement. This will be useful in the future,should learn to know how it works than to know the keyword. Thank you Gagnon.:)

Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2020, 14:18

Take, for example, this code:

for(int i = 2; i <= 100; i++){
   if(i % 2 != 0)
      continue;
   System.out.println(i);
}

This code will output all the even numbers from 2 to 100. If the statement on line 2 is true, when - i - is an odd number, then the continue word instructs the loop to stop that cycle and start a new cycle. This means that line 4 will not be reached whenever - i - is odd and results in only the even numbers being output. It is important to remember that "continue" only works on loops. You can not have it outside of a loop. You can have the keyword in if statements, nested if statements, switch statements, a combination of both, or any other logic gate AS LONG AS they are contained within a loop.

dhanush

Level 30 , Chennai, India

1 September 2020, 14:34

Really! i just ran that code in my pc and understand the difference between the continue, return, break statements. Yeah i know that continue keyword works on loops only.:) But when i use that code(seferi code) in IntelliJ it tells that you don't want continue keyword because it is at the end of the if statement similarly it gives like this and even when i remove that also it works fine i think so.

Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2020, 14:55

The last 2 'continue' statements in his code are unnecessary because the loop is at the end, and it will have the same exact behavior if they are removed. Its sloppy, but expected as learning continues. At least it makes sense and can be improved upon. I came up with some very bizarre and overly complicated methods of doing things when I first started programming that would have had you really scratching your head and asking "Why?".

Seferi

Level 22 , United Federation of Planets

1 September 2020, 19:28

No idea what 'return' does. Gotta read up on that...

Guadalupe Gagnon

Level 37 , Tampa, United States

1 September 2020, 20:33

"return" is the end of a method and will return the running of the code to the calling location or the method (running of the code meaning the line by line path the processing of the code follows). When the method has a return value then a return statement MUST be reached in that method that returns a value that matches the return value of the method. This is true no matter how many branches are in the logic of the method, each branch has to lead to a valid return statement. When a method does not have a return value, meaning its return value is 'void' like the main method or the methods in the above code, it does not need a return statement as the code will naturally return when it finishes executing. When "return" is used in a void method, then the method stops at that point and returns to its calling location. It is similar to "break" in a loop, and just like that "break", "return" would be used behind some if statement and a boolean check. Remember that method declarations MUST have a return value, even void. So, some examples:

// method declared with an int return value
int someMethod1(){
   int x;
   // some code that initializes x and does some processing
   return x; // needs to have a return of an int value
}

// method declared with an ArrayList<Object> return value
ArrayList<Object> someMethod2(){
   return new ArrayList<Object>(); // needs to return an Object ArrayList
}

// a method with no return value, can have a return statement but does not
// need one. If it reaches a return statement then the method ends. Can not
// return any value.
void someMethod3(){
   // some code
   if(/*boolean check*/){
      //if this is true then method just stops, same as "break" in a loop
      return; // no value should be returned, just use "return"
   }
   // some other code if the method didn't return
}