I have two case classes
case class StringCaseClass(argument: String)
case class IntCaseClass(argument: Int)
I want to define a structural type which will match the companion object of both of these
type HasApply1 {
def apply[A, R](argument: A): R
}
This will compile fine, but when I try to use it like this
def method(caseClass: HasApply1) {
// whatever
}
method(StringCaseClass)
I will get a compiler error
found : StringCaseClass.type
required: WithApply1
(which expands to) AnyRef{def apply[A, R](string: A): R}
Is there any way of accomplishing this? If I redefine the structural type to have concrete types for A and R it will compile correctly, but then I lose the flexiblity
The actual type arguments of a generic type are. reference types, wildcards, or. parameterized types (i.e. instantiations of other generic types).
Assigning Generic ParametersBy passing in the type with the <number> code, you are explicitly letting TypeScript know that you want the generic type parameter T of the identity function to be of type number . This will enforce the number type as the argument and the return value.
In order to use a generic type we must provide one type argument per type parameter that was declared for the generic type. The type argument list is a comma separated list that is delimited by angle brackets and follows the type name. The result is a so-called parameterized type.
Multiple parameters You can also use more than one type parameter in generics in Java, you just need to pass specify another type parameter in the angle brackets separated by comma.
@aloiscochard's comment is almost there. What he forgot to mention is that case class companion objects already implement the appropriate FunctionN
trait, so you can simply do this,
scala> case class StringCaseClass(argument: String)
defined class StringCaseClass
scala> case class IntCaseClass(argument: Int)
defined class IntCaseClass
scala> def method[A, R](caseClass: A => R, a: A) = caseClass(a)
method: [A, R](caseClass: A => R, a: A)R
scala> method(StringCaseClass, "foo")
res0: StringCaseClass = StringCaseClass(foo)
scala> method(IntCaseClass, 23)
res1: IntCaseClass = IntCaseClass(23)
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