按姓名呼叫与正常顺序
Call by name vs normal order
我知道这个话题已经讨论过好几次了，但我还是不清楚。
我已经阅读了这个问题 applicative-order/call-by-value and normal-order/call-by-name differences 并且有一些事情我想一劳永逸地澄清：
点名
As normal order, but no reductions are performed inside abstractions. For example λx.(λx.x)x is in normal form according to this strategy, although it contains the redex (λx.x)x.
在按名称调用中，表达式 λx.(λx.x)x 被认为是范式；这是因为“(λx.x)x”被认为是主体（因为 λ 的范围尽可能向右延伸）？那么另一方面，如果我应用正常顺序，结果会是什么？
In call by name, the expression λx.(λx.x)x is said to be in normal form; is this because "(λx.x)x" is considered to be the body (since the scope of λ extends as far as possible to the right)?
是的，你是对的。
And so on the other side, if I apply the normal order, what would be the result?
你在 body 中做了归约：(λx.x)x -> x，所以整个事情归结为恒等函数：
λx.(λx.x)x -> λx.x
为了进一步澄清，让我再做一次，renaming the variables to conform with the Barendregt variable convention：λx.(λx.x)x =α λx.(λy.y)x：
λx.(λy.y)x -> λx.[y := x](y) = λx.x
我知道这个话题已经讨论过好几次了，但我还是不清楚。 我已经阅读了这个问题 applicative-order/call-by-value and normal-order/call-by-name differences 并且有一些事情我想一劳永逸地澄清：
点名
As normal order, but no reductions are performed inside abstractions. For example λx.(λx.x)x is in normal form according to this strategy, although it contains the redex (λx.x)x.
在按名称调用中，表达式 λx.(λx.x)x 被认为是范式；这是因为“(λx.x)x”被认为是主体（因为 λ 的范围尽可能向右延伸）？那么另一方面，如果我应用正常顺序，结果会是什么？
In call by name, the expression λx.(λx.x)x is said to be in normal form; is this because "(λx.x)x" is considered to be the body (since the scope of λ extends as far as possible to the right)?
是的，你是对的。
And so on the other side, if I apply the normal order, what would be the result?
你在 body 中做了归约：(λx.x)x -> x，所以整个事情归结为恒等函数：
λx.(λx.x)x -> λx.x
为了进一步澄清，让我再做一次，renaming the variables to conform with the Barendregt variable convention：λx.(λx.x)x =α λx.(λy.y)x：
λx.(λy.y)x -> λx.[y := x](y) = λx.x