Sorry, but I must ask what that weird numbers mean.

Pytania dotyczące zadania

Rozwiązane

Sorry, but I must ask what that weird numbers mean. long l = (byte) 1234_564_890L; my output is 26. What that lower underline means, and that L? Where I can read about this? I don't want to do this at random (blind hit).

Usuń niepotrzebne operatory tak, aby otrzymać prawidłową odpowiedź: 1234567
long l = (byte)1234_564_890L;
int x = (byte)0b1000_1100_1010;
double m = (byte)110_987_654_6299.123_34;
float f = (byte)l++ + 10 + ++x - (float)m;
l = (long) f / 1000;

Wymagania:

Program powinien wyświetlać tekst na ekranie.
Nie zmieniaj komendy wyświetlającej się na ekranie.
Metoda main() musi zawierać zmienną l typu long.
Metoda main() musi zawierać zmienną x typu int.
Metoda main() musi zawierać zmienną m typu double.
Metoda main() musi zawierać zmienną f typu float.
Nie zmieniaj początkowej wartości zmiennych w trakcie inicjalizacji. Możesz jedynie zmieniać operatory rzutowania.
Program powinien wyświetlać 1234567.

package pl.codegym.task.task10.task1007; /* Zadanie nr 7 z konwersją typu int */ public class Solution { public static void main(String[] args) { long l = (byte) 1234_564_890L; System.out.println("l: " + l ); int x = (byte) 0b1000_1100_1010; System.out.println("x: " + x); double m = (byte) 110_987_654_6299.123_34; System.out.println("m: " + m); float f = (byte) l++ + 10 + ++x - (float) m; System.out.println("f: " + f); l = (long) f / 1000; System.out.println(l); } }

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Guadalupe Gagnon

Poziom 37 , Tampa, United States

19 sierpnia 2021, 18:56

rozwiązanie

Java allows use of underscore in numbers in to break them up, sort of like commas are used. They are pretty much for readability though as these are all identical to the compiler See:

System.out.println(12_34564_8_9_0L);
System.out.println(1234564890L);
System.out.println(1_234_564_890L);

The reason you got 26 is because a byte can only hold 256 unique numbers from -128 to 127. When you go over the individual bytes (the 1's and 0's) flip. Take this code for example:

for(int i = 124; i < 130; i++) {
   System.out.println((byte)i);
}

Run that code and watch what happens when i is greater than 127. If I kept this code looping until for an additional 256 then the bytes would flip again, and so on and so on. If you were counting down instead of up the flip would occur at after -128. The output of 26 is the result of this occurring 1,234,564,890 times

Guadalupe Gagnon

Poziom 37 , Tampa, United States

19 sierpnia 2021, 19:03

The L is just signifying that the number is a long. Java defaults numbers without decimals to 'int', and numbers with decimals to 'decimal'. Also you are going to have to get used to looking things up online, it comes with the territory as a programmer.

Robert

Poziom 20 , Debica, Poland

19 sierpnia 2021, 19:45

I am trying to dig deeply, and there shows I have a question - what I am looking for? I see once something similar in YT Harvard CS50 video but, I must find it and watch again, maybe I find something there. I forgot try to look, discover forward lesson, maybe there I can find something.

Guadalupe Gagnon

Poziom 37 , Tampa, United States

19 sierpnia 2021, 19:54

Are you reading any Java books? If not I really highly recommend one. Codegym's lessons are very abbreviated and skip a lot of useful information. I personally have read "Thinking in Java" and can recommend that. A lot of people also say that "Headfirst Java" is a great book too. Here is what I found on google about your original question: https://zetcode.com/lang/java/datatypes/ It goes over the underscore, longs, data overflows, and a bunch of other concepts.

Robert

Poziom 20 , Debica, Poland

19 sierpnia 2021, 20:19

Yes, I have few. I have that second. But that first I must get. Headfirst Java is different than others. I first time met that kind book, teaching method. Thank you Gaudalupe. Tomorrow I will continue, it is time to sleep for me. 😴

Robert

Poziom 20 , Debica, Poland

22 sierpnia 2021, 09:35

I made it like that. ((1234564890+1)+10+(2250+1)-1)/1000 = 1234567

public class Solution {
    public static void main(String[] args) {

        long l = (byte) 1234_564_890L;

        //System.out.println("l: " + l );
        //System.out.println("l without (byte): " + 1234_564_890L);
        //long y = 1234_564_890L;
        //long yy = y++;
        //System.out.println("l++ without (byte): " + yy);
        //System.out.println();

        int x = 0b1000_1100_1010;

        //System.out.println("x: " + x);
        //System.out.println("x without (byte): " + 0b1000_1100_1010);
        //System.out.println();

        double m = (byte) 110_987_654_6299.123_34;

        //System.out.println("m: " + m);
        //System.out.println("m without (byte): " + 110_987_654_6299.123_34);
        //float z = (float) m;
        //System.out.println("float m: " + z);

        //System.out.println();

        float f = l++ + 10 + ++x - (float) m;

        //System.out.println("f: " + f);
        //System.out.println();

        l = (long) f / 1000;

        System.out.println(l);
        //System.out.println();

        // ((1234564890+1)+10+(2250+1)-1)/1000 = 1234567

        // l = del (byte)
        // x = del (byte)
        // f = del (byte) from l++

    }
}

I have one issue. The value changed at 'l++' is never used.

My variable l is increased by +1, and then is switched to f (line 33), but in my f variable l++ is never used, and never take to count? But there is somehow 1234564890+1, if I throw away l++ my output is equals 2. Where I missed order?

Guadalupe Gagnon

Poziom 37 , Tampa, United States

23 sierpnia 2021, 14:04

It is never used because the ++ post fix (after the variable) increases it by one AFTER the equation. Line 28 processes identically like this:

f = l + 10 + (x+1) - (float) m;
/*then*/ l = l + 1;

I just broke out the increase of 'l +1' to the next line to show you that it processes AFTER the equation. The next line of code, line 33, overwrites the value of 'l' with "(long) f / 1000". This means that the increase of 'l' by 1 is never used. If you removed it the output would be exactly the same. This message is a compiler warning that just lets you know there may be an error in your code because you changed the value of a variable but never used it. This won't prevent code from compiling.