带有关闭的惰性变量 [英] lazy variable with closure

问题描述

在此文章，它说(请参考下面的代码):您必须使用惰性来防止多次创建闭包."

In this article, it says (referencing the code below): "You must use lazy to prevent the closure for being created more than once."

private lazy var variable:SomeClass = {
    let fVariable = SomeClass()
    fVariable.value = 10
    return fVariable
}()

为什么懒惰会阻止多次创建闭包?为什么缺乏惰性会导致它多次评估呢?

Why would lazy prevent the closure from being created more than once? And why would the lack of lazy cause it to evaluate more than once?

推荐答案

您引用的教程代码是这样的:

The tutorial code you quote is this:

private lazy var variable:SomeClass = {
    let fVariable = SomeClass()
    fVariable.value = 10
    return fVariable
}()

与此相反:

private var variable:SomeClass {
    let fVariable = SomeClass()
    fVariable.value = 10
    return fVariable
}

第一个将variable初始化为一个新创建的SomeClass实例，该实例最多一次 (也许没有那么多次).第二个是只读的计算变量，每次读取其值时都会创建一个新的SomeClass实例.

The first one initializes variable to a newly created SomeClass instance, once at most (and maybe not even that many times). The second one is a read-only computed variable and creates a new SomeClass instance every time its value is read.

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