String.equals() method in Java compares the backing arrays of two strings. So if the arrays are of different size same strings would not be equal.
Example:
char[] in = new char[50];
in[0] = 'K';
in[1] = 's';
in[2] = 'h';
in[3] = 'i';
in[4] = 't';
in[5] = 'i';
in[6] = 'z';
String s = new String(in);
System.out.println(s.equals("Kshitiz"));
Output:
false
You should not be creating strings that don't hold what they should hold. Using a hack like trim to deal with bad objects is a bad practice.
Assuming you can't fix the API giving you badly sized char arrays, just count the not null chars before creating the string.
char[] in = new char[50];
in[0] = 'K';
in[1] = 's';
in[2] = 'h';
in[3] = 'i';
in[4] = 't';
in[5] = 'i';
in[6] = 'z';
int length = 0;
while (length<in.length && in[length]!=(char)0) length++;
String s = new String(in, 0, length);
equals() returns false because the two strings are not the same.
If you actually print the original string, then you will see the difference.
public static void main(String[] args) {
char[] in = new char[50];
in[0] = 'K';
in[1] = 's';
in[2] = 'h';
in[3] = 'i';
in[4] = 't';
in[5] = 'i';
in[6] = 'z';
String s = new String(in);
System.out.println(s.equals("Kshitiz"));
System.out.println(s+ "X"); // new line
}
O/P :
false
Kshitiz X
You can see that the two strings are not the same.
Java's Strings are not null-terminated.
If you create Strings from a byte or character buffer, use a constructor that lets you specify the length.
new String(bytes, 0, 7, "ASCII");
new String(characters, 0, 7);
trim() is the way to go if you get char[] from somewhere else.
I added one more code line to your code to show what happen here.
char[] in = new char[50];
in[0] = 'K';
in[1] = 's';
in[2] = 'h';
in[3] = 'i';
in[4] = 't';
in[5] = 'i';
in[6] = 'z';
String s = new String(in);
System.out.println(s.equals("Kshitiz"));
//just use following
System.out.println(s);
Now run this code. you can see s having \u0000 char just after Kshitiz
chararrays are just likecharvalues, they're not convenient and should rarely be used.