Enthuware

Hello,

Does anyone know why
String s = 63 + new Integer(10); will lead to a compile error?

The explanation is: Since none of '+' the operands is a String, the + operator will not generate a String. However, due to auto-unboxing, it will generate an int value of 73.

Thank you!

Since the variable s is of type String, you need a String on the right hand side of the assignment. Now, on the right hand side, you have 63 + new Integer(10) , which does not generate a String. It is an int, which cannot be assigned to s. So the compiler will complain.

For + to generate a String, at least one of the operands must be a String. So, had it been something like 63+"10", the resulting value would have been "6310", a String, which can be assigned to s.

Thank you

Hi Paul,

I have a question regarding the expression:

System.out.println('b'+new Integer(63));

Normally, this will evaluate to 98 + 63 = 161

Why is this accepted as none of the components are strings?
Is it because it is inside of a "println" statement?

Thanks

No, it is accepted because both the sides of the operator are numeric, which makes is the basic usecase for the + operator i.e. addition. Remember that char is also a numeric data type. Char is rot a String.
Also, remember that Integer will be unboxed into an int.

Hi,

Yes, you are right, the Integer is unboxed and then it is just like using an addition between two numeric values.
Thanks

I don't understand why 'b' transfers to 98. Can you provide me with any link where i could read about this topic?

98 is the unicode (and ascii) value of character b. If you use a char as an int, you will get the unicode value of that char. You may read more about it here: https://coderanch.com/t/404152/java/con ... er-integer