嵌套闭包分辨率在方法和属性之间是否不同？ [英] Nested closure resolution different between methods and properties?

问题描述

当闭包的resolveStrategy设置为 DELEGATE_ONLY 或 DELEGATE_FIRST 时，嵌套闭包中方法和属性之间的分辨率不同代表。例如，在下面的示例中， x 解析为 f 的委托（我所期望的），但 keySet（）解析为 g 的委托。

When a closure's resolveStrategy is set to DELEGATE_ONLY or DELEGATE_FIRST, resolution is different in nested closures between methods and properties of the delegate. For example, in the following, x resolves to f's delegate (what I expect), but keySet() resolves to g's delegate.

​def g = {->
    def f = {
        {-> [x, keySet()]}()
    }

    f.resolveStrategy = Closure.DELEGATE_ONLY
    f.delegate = [x: 1, f: 0]

    f()
}      

g.delegate = [x: 0, g: 0]
g()

结果： [1，['x'，'g']]

无嵌套闭包

def g = {->
    def f = {
        [x, keySet()]
    }

    f.resolveStrategy = Closure.DELEGATE_ONLY
    f.delegate = [x: 1, f: 0]

    f()
}      

g.delegate = [x: 0, g: 0]
g()

结果： [1，['x'，'f']]

此行为是否在预期中并记录在某处？

Is this behavior expected and documented somewhere? Is it a bug?

推荐答案

我相信这是一个错误。如果更改 Expando 的地图，其行为将有所不同：

I believe it is a bug. If you change the map for a Expando it behaviors differently:

f = {
  g = {
    { -> keySet() }()
  }

  g.delegate = new Expando(a: 1000, b: 900, c: 800, keySet: { 'g keyset' })
  g.resolveStrategy = Closure.DELEGATE_ONLY
  g()

}

f.delegate = new Expando(a: 90, x: 9, y: 1, keySet: { 'f keyset' })

assert f() == 'g keyset'




f = {
  g = {
    { -> keySet() }()
  }

  g.delegate = [a: 1000, b: 900, c: 800]
  g.resolveStrategy = Closure.DELEGATE_ONLY
  g()

}

f.delegate = [a: 90, x: 9, y: 1]

assert f().toList() == ['a', 'b', 'c'] // fails :-(

也许填写 JIRA ？

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