what is wrong here?

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package com.codegym.task.task04.task0421; /* Jen or Jen? */ import java.io.*; public class Solution { public static void main(String[] args) throws Exception { //write your code here BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in)); String m = bufferedReader.readLine(); String n = bufferedReader.readLine(); int c = m.length(); int d = n.length(); if (m == n && c == d){ System.out.println("The names are identical"); } else if (c ==d ){ System.out.println("he names are the same length"); } } }

Use the keyboard to enter two names. If the names are identical, display "The names are identical".
If the names are different, but they are the same length, display "The names are the same length".
If the names and name lengths are different, don't display anything.

Requirements:

The program should read two lines from the keyboard.
The program should contain System.out.println() or System.out.print().
If the names are identical, display "The names are identical"
If the names are different, but they are the same length, display "The names are the same length"
If the names and name lengths are different, don't display anything.

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Emil

Level 4 , Harnosand, Sweden

25 January 2020, 11:07

You are doing it to complicated. Use only Strings and try to use If(String.equals(String)); on the identical part of your misson. Goodluck Space Soldier!

Guadalupe Gagnon

Level 37 , Tampa, United States

26 November 2019, 19:12

When comparing strings together use the .equals() method or you are actually comparing object addresses. The correct code would be:

if (m.equals(n) && c == d){
//...

The reasons why are a bit convoluted, but I will try to explain a little bit. Because strings values in code are variable length and can hold nearly infinite values (limited by computer resources of course), they use some slick back end programming to help the computer manage its resources more efficiently. When strings have the same value, they literally point to the same place in memory instead of 2 separate locations. When you use the '==' operator on strings, you are checking to see if both strings point to the same location in memory. This, in most cases, won't be an issue because the back end programming handles it very well, but there are situations where the same exact string will not point to the same location. Now the method .equals() on a string variable will check the actual value stored in string and not the address. As an example, run this code:

public class Solution {
    public static void main(String[] args) throws Exception {
        String x = "jen";
        String y = new String("jen"); // this line over rides the backend and creates a new location in memory
        System.out.println(String.format("value of x = '%s' ---- value of y = '%s'",x,y));
        System.out.println(String.format("Does x == y? %s", x==y));
        System.out.println(String.format("Does x.equals(y)? %s", x.equals(y)));
    }
}

The output will show that both x and y store the same value 'jen', but the == operator returns false on them while the .equals() method returns true. If you change the code to remove the new String() so the line is: String y = "jen" and run it, now both the == operator and .equals() method return true. Just remember, if any, to use .equals().