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  2. Java
  3. Data Structures
  4. Java Generics
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Java Generics

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  • yangzhouinter
     + 0 comments
    public class Solution {
    static void printArray(Object[] arr) {
        for (Object o : arr) {
            System.out.println(o.toString());
        }
    }

    public static void main(String[] args) {
        Object[] arr = {1, 2, 3, "Hello", "World"};
        
        printArray(arr);
    }
}

  • MelodyLederman3
     + 0 comments

    This is a great exercise to understand the power of Java generics! Funinexchange Login

  • macohip968
     + 0 comments

    Yeah, the key here is to use a generic method instead of overloading—something like public static void printArray(T[] array) will handle both integers and strings. I learned that the simple approach often saves time, kind of like when I used rapidtestmedellin for quick and confidential health testing—it just worked smoothly without extra complications.

  • sumit_next21
     + 0 comments

    Giving 0 star to this qus.

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        List<Integer> arInt = new ArrayList<>();
        arInt.add(1);
        arInt.add(2);
        arInt.add(3);
        
        List<String> arStr = new ArrayList<>();
        arStr.add("Hello");
        arStr.add("World");
        printArray(arInt);
        printArray(arStr);
    }
    
    public static<T> void printArray(List<T> o){
        o.forEach(a -> System.out.println(a.toString()));
    }

  • julian_n_robbins
     + 0 comments

    In the spirit of the challenge:

        public static <T> void printArray(T[] array) {
        for( T e : array) {
            System.out.println(e);
        }
    }

    public static void main(String[] args) {
        Integer[] number = {1,2,3};
        String[] s = {"Hello", "World"};
        
        printArray(number);
        printArray(s);
    }