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Q.9 from Majji's paper 1

 
Greenhorn
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The following code will give
1: Byte b1 = new Byte("127");
2:
3: if(b1.toString() == b1.toString())
4: System.out.println("True");
5: else
6: System.out.println("False");
A) Compilation error, toString() is not avialable for Byte.
B) Prints "True".
C) Prints "False".
The correct answer is C. I thought it was B because b1.toString() gives the memory address of b1. Can any one help me with this.
Thanx in adv.
 
Ranch Hand
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toString() is a method available to every Object (it's in Object class). Byte class over-rides it - it returns a String object representing the byte. So when you do that if clause - toString is called twice and each time returns a String object. Now those string objects will have the same contents. But they will be separate objects. So using == will return false. Using equals() will return true.
Hope this helps,
Kathy
 
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