在嵌套函数中设置一个可变字段 - deftype - clojure [英] Setting a mutable field in a nested function - deftype - clojure

问题描述

编辑：
发布我的问题的以前版本后，我发现真正的问题是与嵌套函数。

After posting the previous version of my question I discovered that the real problem is with nested functions.

如果我在 deftype 中有一个闭包，我无法更新该闭包内的任何可变字段。

If I have a closure within a deftype I can not update any mutable fields from within that closure.

以下工作：

(deftype Test [^:unsynchronized-mutable x]
    TestInterface
    (perform [this o] (set! x o)))

但这不：

(deftype Test [^:unsynchronized-mutable x]
    TestInterface
    (perform [this o] (fn [] (set! x o)) nil)) ; throws a compiler error about assigning to non-mutable field

有任何方法可以访问领域？执行（set！（.x this）o）导致：

Is there any way to reach up and access the field? Doing (set! (.x this) o) results in:

ClassCastException user.Test可以转换为compile__stub.user.Test user.Test / fn - 152（NO_SOURCE_FILE：3

尝试运行代码时

TestInterface 代码的完整性：

(definterface TestInterface (perform [o]))

推荐答案

考虑不工作的版本，如果它的工作，将返回一个可以逃逸到野外，并从任何地方调用，设置您的不同步的本地和混乱

Consider that the version which does not work would, if it did work, return a closure which could escape into the wild and get called from wherever, setting your unsynchronized local and messing things up terribly.

如果你坚持做这样的事情，你总是可以创建一个接口，使用一个setter为你的可变字段，实现它在你的类型和调用

If you insist on doing this sort of thing, you can always create an interface with a setter for your mutable field, implement it in your type and call the setter wherever you need to set this field.

直接变异（（set！（。-x foo）...））不会工作，因为Clojure类型的可变字段（非同步和易变）是私有的。

Direct mutation ((set! (.-x foo) ...)) won't work, since mutable fields of Clojure types (both unsynchronized and volatile) are private.

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