CODEWITHSUNDEEP
genericskotlin
hkoosha
hkoosha

Reputation: 1146

How to call a function in kotlin with raw types

when a function declares a type parameter:

fun <T> typedFunction(value: T, option: Option<T>) { ... }

how should I call the raw un-typed typedFunction in kotlin?

why?

In java I have:

// This is a method in an external library. I can not change it.
void <T> typedFunction(T t, Option<T> o) { ... }

// This is my code. optionsValues contains many types
// such as Option<Integer>, Option<String>, and ...
Map<Option<?>, ?> m = readAndParseFromConfigFile();
for (Map.Entry<Option<?>, ?> e : m.entrySet()) {
   // The cast does the trick!
   // I know my code is safe, I can tell the compiler to back off.
   typedFunction((Option) e.getKey(), e.getValue());
}

On the call site I'm looping over multiple values whose exact type is unknown (but known to be safe, both parameters conform to same type), hence I have to cast it into a raw type.

How can achieve the same in kotlin?

What IntelliJ does:

This is how IntelliJ converts my code:

val m: Map<Option<*>, *>? = ...
for ((key, value) in m!!) {
    typedFunction<*>(key, value)
    //           ^^^  ERROR!!
}

But it gives an error afterwards: "Projections are not allowed on type arguments of functions and properties"

Upvotes: 3

Views: 1064

Answers (1)

hotkey
hotkey

Reputation: 147951

Since Kotlin does not have raw types and does not provide a star-projection equivalent for function calls, there should be a concrete type for T.

You can make an unchecked cast of the Option<*> argument to Option<Any>, so that T becomes Any:

val m: Map<Option<*>, *>? = ...

for ((key, value) in m!!) {
    @Suppress("unchecked")
    typedFunction(key as Option<Any>, value) // inferred T := Any
}

Upvotes: 2

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