groovy Expando：为什么它考虑局部变量而不是 Expando 属性？
查看此测试代码：
def a = "test"
def expando = new Expando()
expando.a = a
expando.foobar = {a}
expando.a = "test1"
assert expando.foobar() != a
为什么最后一个断言失败？它将“a”视为局部变量，而不是 Expando.a 属性。
感谢您的帮助
looking at this test code:
def a = "test"
def expando = new Expando()
expando.a = a
expando.foobar = {a}
expando.a = "test1"
assert expando.foobar() != a
why the last assertion fail? it considers "a" as the local variable and not as the expando.a properties.
Thanks for help
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也许我错了，但是当您调用expando.foobar()时，它返回分配给foobar的闭包结果。在本例中，它是
a，因此它返回 a 的值：test。expando.foobar()不会调用属性“a”，因为闭包不会查找其委托，除非变量未在范围内定义（在本例中是这样）。编辑：
如果您要执行
expando.foobar = {delegate.a}，则会返回您期望的结果。Perhaps I am mistaken, but when you invoke
expando.foobar(), it returns the result of the closure that was assigned tofoobar. In this case, it isa, so it returns the value of a:test.expando.foobar()does not call the property 'a' because closures do not look for their delegate unless a variable is not defined in scope (and in this case it is).Edit:
If you were to do
expando.foobar = {delegate.a}, that would return the results you are expecting.