Reactor:更新对象的状态 [英] Reactor: Update objects's state

问题描述

给定以下方法:

private Mono<UserProfileUpdate> upsertUserIdentifier(UserProfileUpdate profileUpdate, String id){
    return userIdentifierRepository.findUserIdentifier(id)
            .switchIfEmpty(Mono.defer(() -> {
                profileUpdate.setNewUser(true);
                return createProfileIdentifier(profileUpdate.getType(), id);
            }))
            .map(userIdentifier -> {
                profileUpdate.setProfileId(userIdentifier.getProfileId());
                return profileUpdate;
            });
}

switchIfEmpty 和 map 操作符改变了 profileUpdate 对象.在 switchIfEmpty 运算符中变异是否安全?关于map，如果我理解正确的话，这是不安全的，对象profileUpdate 必须是不可变的，对吧?例如:

switchIfEmpty and map operators mutate the profileUpdate object. Is it safe to mutate in switchIfEmpty operator? Regarding map, if I have understood correctly, this is not safe and object profileUpdate must be immutable, right? eg:

private Mono<UserProfileUpdate> upsertUserIdentifier(UserProfileUpdate profileUpdate, String id){
        return userIdentifierRepository.findUserIdentifier(id)
                .switchIfEmpty(Mono.defer(() -> {
                    profileUpdate.setNewUser(true);
                    return createProfileIdentifier(profileUpdate.getType(), id);
                }))
                .map(userIdentifier -> profileUpdate.withProfileId(userIdentifier.getProfileId()));
    }

在链的后面，另一个方法改变了对象:

Later in the chain, another method mutates the object:

public Mono<UserProfileUpdate> transform(UserProfileUpdate profUpdate) {
        if (profUpdate.isNewUser()) {
            profUpdate.getAttributesToSet().putAll(profUpdate.getAttributesToSim());
        } else if (!profUpdate.getAttributesToSim().isEmpty()) {
            return  userProfileRepository.findUserProfileById(profUpdate.getProfileId())
                    .map(profile -> {
                        profUpdate.getAttributesToSet().putAll(
                                collectMissingAttributes(profUpdate.getAttributesToSim(), profile.getAttributes().keySet()));
                        return profUpdate;
                    });
        }
        return Mono.just(profUpdate);
    }

以上方法调用如下:

  Mono.just(update)
  .flatMap(update -> upsertUserIdentifier(update, id))
  .flatMap(this::transform)

推荐答案

含糊不清的答案，但...视情况而定！

Nebulous answer, but... it depends!

在返回的 Mono 或 Flux 中改变输入参数的危险来自于 Mono 或 Flux 可以多次订阅.在这种情况下，您的手上突然有了共享资源，这可能会导致令人费解的问题.

The danger of mutating an input parameter in the returned Mono or Flux comes from the fact that said Mono or Flux could be subscribed multiple times. In such a case, suddenly you have a shared resource on your hands, and it can lead to puzzling issues.

但是如果有问题的方法是从一个控制良好的上下文中调用的，它可能是安全的.

But if the methods in question are called from a well controlled context, it can be safe.

在您的情况下，flatMap 确保内部发布者仅订阅一次.所以只要你只在这样的 flatMap 中使用这些方法，它们就可以安全地改变它们的输入参数(它保持在 flatmapping 函数的范围内).

In your case, flatMap ensures that the inner publishers are subscribed to only once. So as long as you only use these methods inside such flatMaps, they can safely mutate their input parameter (it is kept in the scope of the flatmapping function).

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