Why i got this error..

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In der Diskussion

i got the error " minimum2" might have been not initlialized" , what i have to do now? I dont get the solution....

Schreibe eine Funktion, welche die kleinste der vier Zahlen berechnet.
Die Funktion min(a, b, c, d) muss die Funktion min(a, b) verwenden (aufrufen).

Hinweis:
Du musst den Rumpf der beiden bestehenden min-Funktionen schreiben.

Anforderungen:

Das Programm muss Text auf dem Bildschirm ausgeben.
Die min-Methoden dürfen keinen Text auf dem Bildschirm anzeigen.
Die Methode main muss min(a, b) und min(a, b, c, d) aufrufen.
Die Methode main muss das Ergebnis der min-Methoden anzeigen. Der Satz soll jedes Mal in einer eigenen Zeile stehen.
Die Methode min(a, b) muss die kleinere Zahlen a und b zurückgeben.
Die Methode min(a, b, c, d) muss die Methode min(a, b) verwenden.
Die Methode min(a, b, c, d) muss die kleinste der Zahlen a, b, c und d zurückgeben.

package de.codegym.task.task02.task0217; /* Die kleinste aus vier Zahlen */ public class Solution { public static int min(int a, int b, int c, int d) { int minimum2; if(min(a,b) > c && d > c){ c = minimum2; }else if(min(a,b) > c && d > d ){ minimum2 = d; } return minimum2; } public static int min(int a, int b) { int minimum1; if (a < b){ minimum1 = a; }else{ minimum1 = b; } return minimum1; } public static void main(String[] args) throws Exception { System.out.println(min(-20, -10)); System.out.println(min(-20, -10, -30, -40)); System.out.println(min(-20, -10, -30, 40)); System.out.println(min(-40, -10, -30, 40)); } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

11 Februar 2021, 14:52

Somethings that may help you out: Variables are unique only to the block of code that they are declared in, from the point they are declared until the end of the block (including sub blocks). This is called their "scope". They do not exist anywhere outside the block. This means that you can use the same variables names over and over in different blocks of code. For example, some quick psuedo code to illustrate:

Solution{ // block 1, the class
    main(String[])( // sub block contained in class block
        var test;
        if(....) { // sub block contained in main
            // var test; !!!ERROR, 'test' already declared on line 3
            var test2; // this is good
        } // end sub 'if' block
        var test2; // this is good because line 6 only exists in the 'if' block which ends on line 7
    } // end main(String[])

   min(int, int)( // sub block 1 block from class block
        var test1;
        var test2;
    } // end min(int, int)

   min(int, int, int, int)  // sub block 1 block from class block
       var test1;
        var test2;
    } // end min(int, int, int, int)
} // end block 1

Guadalupe Gagnon

Level 37 , Tampa, United States

11 Februar 2021, 14:52

Line 3: a variable is declared, test. This variable is unique all the way until the end of the block, which would be line 9. It can be used anywhere between those 2 points. Line 5: You wouldn't be able to make a new variable called 'test' here because line 3 already has a variable called 'test' which is in scope at this point. Line 6: You can make a new variable, I called it test2, here. This variable is declared in the block of code that is the 'if statement. This means it is only usable from the point it is declared until the end of the if statement, which is line 7 Line 8: You can make a variable called test2 here because one does not exists in the block of code at this point because the test2 above has already gone out of scope when the if statement ended Line 9: All variables inside the main method would go out of scope at this point Line 11: A new method is declared. At this point you would have the method paramenters, which are (int, int) (i left variable names out in this example, but in your code they would be a and b) and they go in scope. Lines 12-13: two more variables, test1 and test2, are declared. Both would only be usable at the point they are decaled until the point the block of code ends, which is line 14 Line 14: All variables from line 11 to here go out of scope Lines 16-19: Same as last method, but now you have 2 additional ints (in your code they would be d and c). All variables would go out of scope at line 19 Line 20: No variables would be in scope past here, where the class ends

Guadalupe Gagnon

Level 37 , Tampa, United States

11 Februar 2021, 15:08

Some vocab words: Parameter(s) - These are the values inside parentheses on a method declaration line. So in your code line 35, the main method, is declared with the parameter String[] named args. Line 21 has two int parameters named a and b; and finally line 8 has four int parameters: a, b, c, d Argument(s) - These are passed to a method when called in code and MUST match the parameters the method is calling. So lines 38-41 all call min() methods, the first with the arguments -20 and -10 which match the params of the min(int,int) method; the second with -20, -10, -30, -40 which match the params of the min(int, int, int, int) method... etc Methods are the system that allows code to communicate with other parts of the code, by passing messages with the arguments, and receiving replies with the return values. In this code the messages are being passed between methods, but later you will be expanding this to pass messages between objects. The first message would be sent on line 38, where you call min() and pass the arguments -20 and -10. This sends the message to the min method with 2 parameters, which uses the two arguments and replies back on line 31 with the result of the algorithm in that method, which processes the minimum of those two numbers. The final result being the returned value is passed to the println() method which outputs it to the console. The next few lines do the same, but now with the min(int, int, int, int) method. This has an additional requirement that it must call the min(int, int) method. Don't think of it as anything more than a way to pass a message and receive a response (which will be the minimum of the two numbers used as the argument).

Gellert Varga

Level 23 , Szekesfehervar, Hungary

11 Februar 2021, 10:26

In line 10 you declared the variable "minimum2" but you didn't give any value to it. It is not problem yet. But in line 13 you want to give to "c" the value of "minimum2" what doesn't exist yet. This is the problem what the compiler also said: "variable minimum2 might not have been initialized". Not initialized means it doesn't have any starter value yet.

Christopher Berg

Level 7 , Germany

11 Februar 2021, 10:37

okay thank you!, i think i understand that,so how can i give the value c to minimum2 ? but in the lesson before, where i have to write a programm, that give the smallest number from 2 different numbers, i did it like this ; public static int min(int a, int b) { int m2; if(a < b){ m2 = a; }else{ m2 = b; } return m2; and this going green.. 🤷‍♀️

Gellert Varga

Level 23 , Szekesfehervar, Hungary

11 Februar 2021, 10:46

Line 16 is OK. I think you want to do in line 13 the same just with variable "c" : minimum2 = c;

Christopher Berg

Level 7 , Germany

11 Februar 2021, 12:43

ok done. now i got the same error in line 18 by statement return minimum2; Sorry for all the questions, i am rly confused with that...

Gellert Varga

Level 23 , Szekesfehervar, Hungary

11 Februar 2021, 13:12

Questions: No problem!:) Line 18, return minimum2; // maybe your { if / else if } conditions may have omitted some of the possibilities, and yet the variable minimum2 did not receive any value. Try to give it a starting value of 0 when declaring it at the very beginning.