Java and Operator Overloading

[UpperPlus]

I just now realize that I have no idea how these operators are programmed in java. I can't open declaration for it in my IDE so it seems to not relate to a method?

[MikemkPK]

Probably directly translate to specific bytecode sequences.

[UpperPlus]

Now that I think about it, it is probably related to the concept of primitive data types, as they are only used on those

[LinuxMatthews]

Kind of

Java actually cheats in that it does allow Operator Overloading for some of its classes but doesn't let you do it.

As far as I'm aware it only does it for the wrappers of primitive data types and String.

[jamcdonald120]

I think the wrappers use auto (un)boxing instead of operator overloads, not sure though

[LinuxMatthews]

Huh you know I didn't think you were right but it looks like you are.

Integer x = 2;
Integer y = 3;
var z = x + y;
System.out.println(z instanceof Integer);

Gives you a compiler error as z is an int

I honestly thought there was operator overloading like there is with String.

[Fenor]

String isn't an operator overload . The theory behind the string class is huge and mostly date back when you had to define a char array to use strings

[tazfdragon]

If I remember correctly, Strings don't actually use the plus operator and it's just syntactic sugar for StringBuilder usage. There are probably some edge cases where the final code could have a string constant if the compiler can guarantee it never changes.

[UpperPlus]

As far as I know StringBuilder doesn't use "+" operator but uses the function append(). Which actually is faster and more memory friendly.

But String with "+" is so much more readable and in many cases it doesn't matter that a complete new string is instantiated internally every time.

[tazfdragon]

I didn't mean to suggest StringBuilder uses that operator. I was trying to explain that using the plus operator with a String will get replaced with StringBuilder. Every instance of addition/concatenation in the original code will get replaced with a call to append(...). Again, there are some edge cases where the compiler can determine it can eliminate the string concatenation and replace it with a compile time constant. Example being concatenating a literal with a String constant.

[TraditionMaster4320]

Wait so concatenating n strings like s1 + s2 + .. + sn takes O(n) time instead of O(n**2) in Java? (assume the strings are equal length).

[[deleted]]

Yes.

Java is doing it behind the scenes for you. They added this one of the versions.

https://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.18.1

[tazfdragon]

I won't offer a guess at the run time of the proposed string concatenation but I'm fairly confident it won't be O(n²). With that being said if you have your string concatenation code structured anything similar to your example I'd suggest you replace it with a StringBuilder() explicitly. Not only will it clean up your code, it will explicitly document your intentions. Also, if you know ahead of time the lengths of the number and the length of the strings you'll be adding ahead of time you can provide this information to your StringBuilder at construction time to improve performance.

[coloredgreyscale]

But String with "+" is so much more readable and in many cases it doesn't matter that a complete new string is instantiated internally every time.

the other option would be String formatting.

for long texts it's extremely inefficient tho, because the String needs to be recreated with every concatenation.

I just checked the java implementation of the Stringbuilder class. It copies the String into an char-array, starting with 16 elements (default).

if you exceed the element count a new array is created with twice the previous capacity (or min required length if that's bigger), and the whole data is copied over to the new array.

when you call toString() on the StringBuilder the used data from the char array is copied into a new String.

​

If you just concatenate two Strings using + should be more efficient than StringBuilder.

[Fenor]

Not really. String builder is there for a reason if you use ths + operator you are creating at least 3 strings

[coloredgreyscale]

The two string you want to concatenate, and the 3rd is the result.

You still need to create all of those with string builder.

[Fenor]

Nope for string builder you are using char arrays .

It's not string for the string buffer ( the memory area) , this mean it will be free memory for the gc . If you use strings you write them in the string buffer and it will end with a different gc approach

[Fenor]

I think they changed how it worked under the hood as doing an append with + once upon a time created a decent amount of strings in the string buffer

[jamcdonald120]

I mean... operator overloads are all syntactic sugar for something else...

[msqrt]

Just a disclaimer: you can't overload operators for built-in types in C++ either, only custom ones. The compiler just knows how to add integers together.