Kotlin - Why do we have to explicit type parameter(s) for generic method? - kotlin

Question

I'm working on extension method like this:
infix fun <T> T.isNullOr(other: T): Boolean {
if (this == null) return true
return this == other
}
and I'm trying to use this method like this.
val thisShouldWork = true isNullOr true // this is true
val thisShouldNotWork = true isNullOr 0 // No compilation errors?
I expected compilation error because type parameter is automatically set to Boolean for isNullOr but it wasn't. What's happening?
am I misunderstanding about it?
in C#, same code working well as I expected.
static bool IsNullOr<T>(this T t, T other) {
if (t == null) return true;
return Equals(t, other);
}
bool howAboutThis = 0.IsNullOr(0);
bool andThis = 0.IsNullOr(false); // error - cannot detect type parameter for this

Answer 1

Here, val thisShouldNotWork = true isNullOr 0 is equal to val thisShouldNotWork: Boolean = true.isNullOr<Any>(0). Type parameter as inferred as the closest parent.
And function's return type is based on logical expression evaluation: this == other. Let's see == function declaration: public open operator fun equals(other: Any?): Boolean. It receives Any?.
Type parameter in this function has nothing to do with Boolean.

Answer 2

Just remember that generic type information is erased at runtime and whenever you try to put something into a method that accepts generics, then the common denominator is assumed, e.g.:
listOf("one", 123) // -> assumes T:Any and therefore gives List<Any>
Now for your example that would mean "one".isNullOr(123) both become Any.
As a sidenote however, if you declare a specific type (e.g. List<String>) as shown next, it will not work to assign a different type to it:
val test : List<String> = listOf(123) // this will not work
It is already known at compile time that the given int can't become a string. This sample however doesn't help you as you do not return that generic type. If your method just looked a bit different, e.g. would have a generic type as return value, it might easily have worked out similar to the List-sample before.
So to fix your sample you need to specify the type which will basically make the infix obsolete, e.g. the following will work as you expect:
val someString : String? = TODO()
val works = someString.isNullOr<String?>("other")
val doesntWork = someString.isNullOr<Int?>(123) // does not nor does:
val doesntWorkToo = someString.isNullOr<String?>(123)
Note that for what you've shown some standard functionality might help you (but not eliminate that specific problem), i.e. using the ?: (elvis operator) with a ?.let:
val someVal : String? = "someString given from somewhere"
val thisWorks = someVal?.let {
it == "some other string to compare"
} ?: true /* which basically means it was null */
val thisWillNot = someVal?.let {
it == 123 // compile error (funny enough: it.equals(123) would work ;-)
} ?: true /* it is null */

Answer 3

I think in this case the generics don't really matter. You only call equals in the method, which you can do on any type. It's basically the same as:
infix fun Any.isNullOr(other: Any): Boolean {
return this == other
}
It compiles without problems because you can always call equals with anything: other: Any?

Answer 4

Thank for answers. I think there is no way to prevent this at compilation level, so I decided to check type for other.
inline infix fun <reified T> T.isNullOr(other: T): Boolean {
if (this == null) return true
if (other !is T) return false
return this == other
}

Answer 5

If you really want to prevent it, you can:
class IsNullOr<T>(val x: T) {
operator fun invoke(other: T): Boolean {
if (x == null) return true
return x == other
}
}
fun <T> T.isNullOr() = IsNullOr(this)
fun main(args: Array<String>) {
val thisShouldWork = true.isNullOr()(true) // compiles
val thisShouldNotWork = true.isNullOr()(0) // doesn't compile
}
This makes type inference depend only on the receiver of isNullOr. If vals could be generic, you'd even keep the original syntax (but they can't).