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Witam, skoro rozwiązanie to 4x System.out.println(s) To skąd wiadomo że "s" zawiera dany String skoro nie widzę zadeklarowanej zmiennej?

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Guadalupe Gagnon

Poziom 37 , Tampa, United States

25 listopada 2022, 05:49

It's a variable. It is assigned a value when the method is actually called in other parts of code. That is what methods are, reusable pieces of code. If you hardcoded a value then it would reduce the methods usefulness. Consider having 10 methods to print the values 0-9, or one method that prints something and then call it 10 times to print the same: So you could write code like this to print the numbers as described above:

public static void main(String[] args) {
   print0();
   print1();
   print2();
   print3();
   print4();
   print5();
   print6();
   print7();
   print8();
   print9();
}

void print0() {
   System.out.println(0);
}

void print1() {
   System.out.println(1);
}

void print2() {
   System.out.println(2);
}

void print3() {
   System.out.println(3);
}

void print4() {
   System.out.println(4);
}

void print5() {
   System.out.println(5);
}

void print6() {
   System.out.println(6);
}

void print7() {
   System.out.println(7);
}

void print8() {
   System.out.println(8);
}

void print9() {
   System.out.println(9);
}

Guadalupe Gagnon

Poziom 37 , Tampa, United States

25 listopada 2022, 05:51

or you could use a general method (as it is supposed to be) to do the same with WAY less typing:

public static void main(String[] args) {
   print(0);
   print(1);
   print(2);
   print(3);
   print(4);
   print(5);
   print(6);
   print(7);
   print(8);
   print(9);
}

void print(int num) {
   System.out.println(num);
}