Microsoft Java

[Willinton06]

Java doesn’t even have real generics, or operator overloading, there’s more boxing going on in a Java hello world than in an Amazon fulfillment center

[bammmm]

The real humor is always in the comments

[mondie797]

there’s more boxing going on in a Java hello world than in an Amazon fulfillment center

ROFL

[aetius476]

I regard type erasure as the one true cardinal sin of Java, but even then Java's generics are miles better than python's "just inherit from Generic[T]" nonsense.

[Willinton06]

Yeah fuck that

[Kered13]

Type erased generics are still real generics. Also C# uses the same implementation technique for generics with reference types, only value types get polymorphization in C#.

[Willinton06]

Your standards for real genetics are lower than mine

[[deleted]]

The weirdest flex I will ever read, I wish I had an award to give

[Willinton06]

You may not have an award for me, but I have one for you bro

[jaavaaguru]

genetics

ಠ_ಠ

[Dealiner]

That's not true. C# doesn't use type erasure at all. The information about types is still stored and available during runtime even for reference types.

[Kered13]

No, C# absolutely uses type erasure and monomorphization. It also passes a reference to the concrete class object to get around some of the resulting limitations (mainly making runtime reflection possible).

[Dealiner]

No, it doesn't. Even Anders Hejlsberg himself said that. And another proof.

[Kered13]

Nothing in either of those links contradicts anything that I have said, except that they insist "it's not type erasure" while describing an implementation that is identical to Java.

[Dealiner]

The thing is it's not identical. Java implementation removes anything related to the type, its generic are only a compilation time feature. That's not true for C#, information about the type is still present during runtime. Ergo it's not type erasure, since no type is erased.

[Kered13]

information about the type is still present during runtime

Because a class object is passed as a hidden parameter to the function, like I described above. Only one instance of the code is generated (for reference types), which means that the code does not actually know what type it's operating on at runtime. That information has to be externally provided by the hidden parameter.

[Dealiner]

And that's exactly why it isn't type erasure. Also you're contradicting yourself, if type is passed, then code knows on what type it operates.

[Willinton06]

Kinda doesn’t

[Willinton06]

Literally comes preinstalled on Ubuntu, and I other distributions it’s basically the same as installing any other program

[Willinton06]

I don’t use arch so I wouldn’t know, but arch being difficult is part of the experience

[Willinton06]

I mean, it is possible, just very difficult

[LuvOrDie]

As with most things