Unable to match generic types in method parameters

Question

I don't know if the title is understandable but I have this:

 public class Product { public Integer getId() {...} public String getName() {...} } public <T,V> static void method(Function<T, V> f, V value) {...}

and I want to become a compile error when:

 method(Product::getId, "some String"); // id is not String method(Product::getName, 123); // name is not Integer

But the compiler interpret V as:

java Serializable & Comparable<? extends Serializable & Comparable<?>>

It compiles but depending on how "method" is implemented, you get an Exception at runtime or it just works wrongly.

How can I instruct the compiler to match the wanted data type? and I don't want to write a "method" for every possible V type.

Thanks!

@AndrewVershinin not really, because they suggest to implement the method like Object.equals but it happens at runtime and I want it at compiling time. But thanks! — César Mora
– César Mora, Commented Mar 1, 2021 at 7:20

Lino · Accepted Answer · 2021-02-26 09:49:28Z

You can enforce the generics by splitting them up accross two method calls, for that you'd need a new class:

public class Value<T, V> { private final Function<T, V> f; public Value(Function<T, V> f) { this.f= f; } public void with(V value) { // move your code from method() into here } }

And then change method() to something like this:

public static <T, V> Value<T, V> method(Function<T, V> f) { return new Value<>(f); }

Then you can use it like this:

method(Product::getId).with("123"); // compiler error method(Product::getName).with(123); // compiler error method(Product::getId).with(123); // no error method(Product::getName).with("123"); // no error

An extra class and an extra method callback but it does the trick. Thanks!

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