I've noticed this peculiarity: ```val a: Int = foo() val b: Int? = bar() if (a < b!!) {...} else ...

# getting-started

[Klitos Kyriacou]

I've noticed this peculiarity:

val a: Int = foo()
val b: Int? = bar()
if (a < b!!) {...}
else if (a == b) {...} 
else if (a == b + 1) {...}

In the case of a == b it calls Int.equals(Any?) so it boxes a and then compares it to the boxed b. But for a == b + 1 it smart-casts b as Int. Why doesn't it smart-cast b in the first case?

[Gleb Minaev]

Compiler assumes that if unsafe cast b!! did not throw any exception then b is Int. Thus, it smarcasts b in a == b + 1 in

if (a < b!!) {} else if (a == b + 1) {}

in the same way as it smartcasts b in a < b in

if (a == b!! + 1) else if (a < b) {}

[Klitos Kyriacou]

Yes, but in that case it has to smart cast b because you can can't do the + and < operations on Int? but they can be done on Int and so the smart cast is necessary. But my question is why does it not smart cast b in the expression a == b? It can but chooses not to, and prefers to box a instead of comparing primitives.

[Gasan]

@Klitos Kyriacou maybe because RHS operand of + in (b + 1) is an int, the compiler converts b to int. So the resulting value has the type int. So now you have a == <some int> , so the compiler converts a to int

[Klitos Kyriacou]

No, a is already an Int.

[Gasan]

@Klitos Kyriacou what is the type of b + 1?

[Klitos Kyriacou]

b + 1 is an Int.

[Klitos Kyriacou]

Actually, looking at the generated bytecode, I think I was mistaken about what the expression a == b does. It's just IntelliJ IDEA says b is not smart-cast, but it seems the compiler does indeed smart-cast it.