CODEWITHSUNDEEP
javalambdatype-conversionjava-8overloading
Andremoniy
Andremoniy

Reputation: 34900

Implicit type conversion for lambda expression

Consider the following piece of class:

public void method() {
    test(() -> { });
}

void test(Runnable a) {
    System.out.println("Test 1");
}

void test(A a) {
    System.out.println("Test 2");
}

interface A extends Runnable {

}

Invoking method method() will lead to Test 2 output. This means, that lambda expression () -> { } was implicitly converted to A. Why?

Upvotes: 3

Views: 422

Answers (1)

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 279940

It's the same standard rule applied to all overloads. Java will choose the most specific applicable method.

Both methods accept an argument that is of a functional interface type. The lambda expression

() -> { }

is convertible to both those types. A is a subclass of Runnable and is therefore more specific. The method with a parameter type of A therefore gets chosen.

Upvotes: 7

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