CodeGym/Pomoc z zadaniami Java/I have an error when my input is for example "q1"

Poziom 17

Debica

27.12.2022
371widoki
16komentarze

I have an error when my input is for example "q1"

Pytania dotyczące zadania

Różne metody dla różnych typów

Podstawy Java, Poziom 5, Lekcja 12

Rozwiązane

Hello, I tried to catch it, but it not work. Finally, I don't know, this has influence to fail the last point.

1. Odczytuj dane z konsoli do momentu wprowadzenia słowa "exit".
2. Dla każdej wartości (za wyjątkiem "exit"), wywołaj metodę print. Jeśli wartość:
2.1. zawiera kropkę („.”), wówczas wywołaj metodę print dla Double;
2.2. jest większa od zera, ale mniejsza niż 128, wówczas wywołaj metodę print dla short;
2.3. mniejsza lub równa zero lub większa lub równa 128, wówczas wywołaj metodę print dla Integer;
2.4. w przeciwnym razie wywołaj metodę print dla String.

Wymagania:

Program musi odczytywać dane z klawiatury.
Program powinien zakończyć odczytywanie danych z klawiatury po wprowadzeniu "exit".
Jeżeli wprowadzony ciąg zawiera kropkę („.”) i można go poprawnie przekonwertować na typ Double, to należy wywołać metodę print(Double value).
Jeżeli wprowadzony ciąg można poprawnie przekonwertować na typ short, a otrzymana liczba jest większa niż 0, lecz mniejsza niż 128, to należy wywołać metodę print(short value).
Jeżeli wprowadzony ciąg można poprawnie przekonwertować na typ Integer i otrzymana liczba jest mniejsza lub równa 0 lub większa lub równa 128, to należy wywołać metodę print(Integer value).
We wszystkich innych przypadkach należy wywołać metodę print(String value).

package pl.codegym.task.task15.task1519; import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; import java.util.ArrayList; /* Różne metody dla różnych typów */ public class Solution { public static void main(String[] args) throws IOException { //tutaj wpisz swój kod BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String keyInput; do { keyInput = reader.readLine(); int countDot = 0; boolean isNum = false; boolean isAlpha = false; for (char c : keyInput.toCharArray()) { if (c == '.') { countDot++; } if (Character.isDigit(c)) { isNum = true; } if (!Character.isDigit(c)) { isAlpha = true; } } if (isNum == true) { // req. 3. if (countDot == 1 && isNum == true) { print(Double.parseDouble(keyInput)); } else if (countDot > 1 && isNum == true) { print(keyInput); } // req. 4. else if (isNum == true && Integer.parseInt(keyInput) > 0 && Integer.parseInt(keyInput) < 128) { print(Short.parseShort(keyInput)); } //req. 5. else print(Integer.parseInt(keyInput)); } // where input is with letters and numbers example "q1" else if (isNum == true && isAlpha == true) { // here I try to fix a problem with input "q1" value, but it didn't work. print(keyInput); } else print(keyInput); } while (!keyInput.equals("exit")); } public static void print(Double value) { System.out.println("To jest Double. Value: " + value); } public static void print(String value) { System.out.println("To jest String. Value: " + value); } public static void print(short value) { System.out.println("To jest liczba całkowita typu short. Value: " + value); } public static void print(Integer value) { System.out.println("To jest liczba całkowita typu Integer. Value: " + value); } }

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Guadalupe Gagnon

Poziom 37 , Tampa, United States

28 grudnia 2022, 16:17

przydatny

Tip: The boolean check part of a loop requires that an expression within it evaluates to true or false. The expression itself can compound and be quite elaborate as long as it ends in true/false. For example, you could read keyboard input, assign it to a variable, and finally check if that value is equal to a condition value. Either of these examples work:

while( !(variable = reader.readline()).equals("exit") ) { }

for(String variable; (variable = reader.readline()).equals("exit") ; ) { }

I will explain how I solved this task and the logic that I used. I always start each task by trying to logically define the parameters set by the conditions. In this task we have 5 possible scenarios with each keyboard entry: 1) "exit" is entered 2) a String is entered 3) a double is entered 4) an int is entered where var > 0 AND var < 128 5) and finally anything else not fitting the above descriptions A number for this task can be defined as optionally starting with a hyphen, containing ONLY the digits 0-9, and optionally containing 1, and only 1, period. step1: So dealing with #1 is easy. If the input equals "exit" then end the program. step2: Next I would check each char in the input. - negative numbers start with a hyphen, if the first char is a hyphen you can continue to next char - If the char is not a digit or a period then you could send directly to print() and read next input - If a period occurs more than once you could send directly to print(String) and read next input - If, by the end, the loop isn't stopped then a number has occurred step3: if the number contains a period then convert to double, send double to print(), and read next input step4: convert to int. If the number int is greater than 0 AND less than 128 then send int to print(short) and read next line Step 5: send int to print() and read next line

Guadalupe Gagnon

Poziom 37 , Tampa, United States

28 grudnia 2022, 16:18

(By "read next line" I mean stop the current loop cycle and continue a new loop cycle) When I did this task, years ago, I didn't use a regular expression (regex) because I hardly understood the syntax for that. If you are good with them then determining a number or a String could be one line instead of having to loop each char.

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 21:07

After Thomas suggestion I do this.

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

    String keyInput;

    while (true) {
        keyInput = reader.readLine();

        //req. 1.
        if (keyInput.equals("exit")) {
            break;
        }

        int countDot = 0;
        for (char c : keyInput.toCharArray()) {
            if (c == 46) {
                countDot++;
            }
        }

        if (isDouble(keyInput)) {

            if (isNum(keyInput)) {
                if (Integer.parseInt(keyInput) > 0 && Integer.parseInt(keyInput) < 128) {
                    print(Short.parseShort(keyInput));
                } else print(Integer.parseInt(keyInput));
            } else print(Double.parseDouble(keyInput));

        } else print(keyInput);

    } //end while loop
} // end main

public static boolean isNum(String s) {

    for (char c : s.toCharArray()) {
        if (c < 48 || c > 57) {
            return false;
        }
    }
    return true;
}

public static boolean isDouble(String s) {
    try {
        Double.parseDouble(s);
    } catch (NumberFormatException e) {
        return false;
    }
    return true;
}

And as result it gives me that.

1
To jest liczba całkowita typu short. Value: 1
1.0
To jest Double. Value: 1.0
-1.0
To jest Double. Value: -1.0
100
To jest liczba całkowita typu short. Value: 100
10000
To jest liczba całkowita typu Integer. Value: 10000
q1
To jest String. Value: q1
2q
To jest String. Value: 2q
qwerty
To jest String. Value: qwerty
-q1
To jest String. Value: -q1
-qwerty
To jest String. Value: -qwerty
.q
To jest String. Value: .q
.200
To jest Double. Value: 0.2

Seems to be working, but no enough.

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 21:10

Because 3 and 5 req is not complete.

Guadalupe Gagnon

Poziom 37 , Tampa, United States

28 grudnia 2022, 21:26

negative integers are not being considered here, such as -5 or -4203.

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 21:30

Then the negative number must be a string, in result?

Guadalupe Gagnon

Poziom 37 , Tampa, United States

28 grudnia 2022, 21:45

If I entered -453 it should output: To jest liczba całkowita typu Integer. Value: -453

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 22:03

Ok got it, the short value is only for 1-127 range, rest numbers are Integer, if they do not have one dot (double). Integer should be for (below or equal, zero range) <= 0 to (+numbers) >=128

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 23:27

Thanks guys, I focused on that what I'm done to this moment. If I wanted to go a way, “every single digit” probably had to rearrange all the code, and I was close to do this. Not bad 42 attemps. 50 more left to throw screen by the window. Not bad. Not bad.🥳

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 00:00

I think, I fix that, but 6 req. still fail.

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

String keyInput;

do {
    keyInput = reader.readLine();

    int countDot = 0;
    boolean isNum = false;

    for (char c : keyInput.toCharArray()) {
        if (c == '.') {
            countDot++;
        }
        if (Character.isDigit(c)) {
            isNum = true;
        }
    }

    try {
        if (isNum == true) {
            // req. 3.
            if (countDot == 1 && isNum == true) {
                print(Double.parseDouble(keyInput));
            }

            else if (countDot > 1 && isNum == true) {
                print(keyInput);
            }

            // req. 4.
            else if (isNum == true && Integer.parseInt(keyInput) > 0 && Integer.parseInt(keyInput) < 128) {
                print(Short.parseShort(keyInput));
            }

            //req. 5.
            else print(Integer.parseInt(keyInput));
        }
    } catch (NumberFormatException e) {
        print(keyInput);
    }

} while (!keyInput.equals("exit"));

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 00:33

I developed req. 5. in line 37 I'm switched loop do-while to while, and still nothing.

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

    String keyInput;

    while (true) {
        keyInput = reader.readLine();

        //req. 1.
        if(keyInput.equals("exit")) {
            break;
        }

        int countDot = 0;
        boolean isNum = false;

        for (char c : keyInput.toCharArray()) {
            if (c == '.') {
                countDot++;
            }
            if (Character.isDigit(c)) {
                isNum = true;
            }
        }

        try {
            if (isNum == true) {
                // req. 3.
                if (countDot == 1 && isNum == true) {
                    print(Double.parseDouble(keyInput));
                }

                else if (countDot > 1 && isNum == true) {
                    print(keyInput);
                }

                // req. 4.
                else if (isNum == true && Integer.parseInt(keyInput) > 0 && Integer.parseInt(keyInput) < 128) {
                    print(Short.parseShort(keyInput));
                }

                //req. 5.
                else if (isNum == true && Integer.parseInt(keyInput) <= 0 || Integer.parseInt(keyInput) >= 128) {
                    print(Integer.parseInt(keyInput));
                }
            }
        } catch (NumberFormatException e) {
            print(keyInput);
        }
    }
}

Thomas

Poziom 41 , Bayreuth, Germany

28 grudnia 2022, 07:14

Your isNum boolean does not contain the correct result in all cases. As soon as there is one digit you consider the entire string to be a number. However you should do it the other way round. If one char is not a number then it is a string. And even if that check is wrong you keep it in all your tries but extend your approach with a try-catch check. I'd suggest to go with one. a) try-catch - Inside parse the String to a double. - If that fails, it is a string. - If not, do some more checks... does the original string contain a dot -> string, then check the rangs... b) assign a correct result to your isNum boolean and go from there, like (you may want to enhance it as it does not check for +, -)

public static boolean isDouble(String s) {
    int dotCount = 0;

    for (char c : s.toCharArray()) {
        if (c == '.')
            dotCount++;
        else if (c < '0' || c > '9')
            return false;
    }

    return dotCount <= 1;
}

Now you can do the same as above with try-catch... check if it is a double... no, then print the string yes, check if the original string contains a dot -> print double no, check ranges...

Guadalupe Gagnon

Poziom 37 , Tampa, United States

28 grudnia 2022, 16:32

Along with what Thomas said, you are also forgetting that negative numbers can be entered. In this case the input would start with a hyphen. You would then still need to determine if all other "number" parameters where satisfied.

Thomas

Poziom 41 , Bayreuth, Germany

28 grudnia 2022, 17:09

Hi Lupe, how have you been ;) The +- problem I've mentioned above also writing a regex for valid doubles is not that easy. I tried once and I had to require a digit before the dot. Otherwise smth like that would be possible "." or "-" or "+."

public static boolean isDouble(String s) {
    return s.matches("^[+-]?\\d+\\.?\\d*$");
}

Robert

Poziom 17 , Debica, Poland

28 grudnia 2022, 20:45

Thomas if, something like this way, maybe I'm a close.

public static boolean isNum(String s) {

        for (char c : s.toCharArray()) {
            if (c < 48 || c > 57) {
                return false;
            }
        }
        return true;
    }

    public static boolean isDouble(String s) {
        try {
            Double.parseDouble(s);
        } catch (NumberFormatException e) {
            return false;
        }
        return true;
    }

Thomas

Poziom 41 , Bayreuth, Germany

28 grudnia 2022, 22:02

rozwiązanie

if you use try-catch, then you do not need anything else. See, if you parse a string using parseDouble and that string is no valid double, then an exception is thrown and you end in the catch block. That means again, you can call the print(String) method (because you know it has to be a string)

try {
    double num = Double.parseDouble(s);
    // if we reach this line, then parsing s to a double worked and we have somthing like 1, 0.1, .3, -44...
    // you can do more checks now as written above...
} catch (NumberFormatException e) {
    // here we know s can not be parsed to a double -> it's a string -> print(s);
}

If you do not want to go this way, then you need to code a more sophisticated way to validate if a string is a number and not just if each char is 0 - 9 - you need to check if it starts with a + or - (allowed), but +/- must not be in the middle of the string - if there is at least one number - if there is a dot and if so if it is just one