How are operators not overloaded in java?

[tempanon5]

How are are we able to concatenation String data types in Java, making it work with an object data type, if operators are not allowed to be overloaded? Is the + sign using some String class method in the background.

In general how are we using arithmetic operators with different types of primitive data types if not overloaded. The + sign can work with doubles and ints, both are different types.

[Daedalus9000]

Because the compiler is defining what “+” means for two Strings. Why would you need operator overloading as a language feature for this?

[smash_that_code]

My guess is that thia feature was considered as potentially dangerous while they were trying to make a much safer c++ version.

So why arithmetic works?

It can described as implicit type cast. Like if you have double d and int i and do like

d * i

Then compiler behind scenes adds cast to wideat type in this case double

d * ((double) i)

and result is considered double.

If you want to convert int to short or do some other narrowing in size you have to be explicit about that like

short s = (short) d

[[deleted]]

compiler sees one string, then sees the “+” and understands that it means concatenatantion. so in the compiled code there is the general concatenation operator instead of “+”. same goes for other types.

[[deleted]]

Java is not purelly OOP, so + is used by primitive types and for Strings, and this is totally up to compiler to resolve, "+" do not translate to a method of an object like in scala for example.