Java 8中Optional的flatMap方法签名合理性疑问

Great question—let's break this down step by step, looking at generics behavior, Optional's design, and why the official signature is the way it is.

First, a quick recap of the original method signature from Oracle's documentation:

<U> Optional<U> flatMap(Function<? super T, Optional<U>> mapper)

1. Would Function<? super T, Optional<? extends U>> work? #

In short: it’s technically possible, but it would require changing flatMap’s return type to Optional<? extends U>—and that tradeoff isn’t worth it for the standard API. Here’s the breakdown:

• Java generics are invariant, meaning Optional<Integer> is not a subtype of Optional<Number> even though Integer extends Number. If your mapper returns Optional<Integer> (which fits Optional<? extends Number> when U=Number), you can’t directly assign that to an Optional<Number> without an unchecked cast.

• The original signature enforces the mapper returns exactly Optional<U> so flatMap can hand back a concrete Optional<U>—no wildcards, no casting headaches for callers. This makes the API predictable and easy to use in most everyday cases.

If we modified the signature to accept Optional<? extends U> and return Optional<? extends U>, it would be type-safe, but it would force developers to deal with wildcard types, adding unnecessary complexity for minimal practical gain.

2. What about Function<? super T, ? extends Optional<? extends U>>? #

This is a non-starter. The key detail here is that Optional is a final class in Java—there are no subclasses of Optional allowed. That means ? extends Optional<? extends U> is exactly equivalent to Optional<? extends U>. Adding this extra wildcard doesn’t give you any additional flexibility; it just makes the signature more verbose for no reason.

A Quick Note on the Contravariant Input (? super T) #

You noticed the input uses contravariance (? super T), which is intentional. This lets you pass a mapper that accepts a supertype of T. For example, a Function<Object, Optional<String>> works perfectly with an Optional<String>, since Object is a supertype of String. This adds flexibility without sacrificing type safety.

Final Takeaways #

• Function<? super T, Optional<? extends U>>: Technically feasible but not ideal for the standard API—prioritizing simplicity and concrete types over wildcard flexibility was the right call for Oracle’s design.
• Function<? super T, ? extends Optional<? extends U>>: Completely unnecessary, as Optional can’t be subclassed. This is just a more verbose version of the first option.

内容的提问来源于stack exchange，提问作者huoenter