It does allow an empty Nil list, however.
不过它确实允许一个空的nil列表。
It calls the primary constructor with Nil (which represents an empty List object).
它调用主构造函数，并且传入Nil (Nil表示了一个空的List对象)。
(It does allow an empty Nil list, however.)
(不过它确实允许一个空的nil列表。)
My standard reply was no, I'm not afraid, because I'm an analyst and the statistical probability of having another such attack is close to nil.
我的回答一直是否定的。我是一名分析师，从数据角度来看，发生类似袭击时间的可能性近乎为零。
Asking an object if it is of the nil type (via the nil? Method call) returns a Boolean.
询问一个对象是否是nil类型(通过nil ？方法调用)会返回一个布尔值。
If you call cons on nil, Lisp treats nil as an empty list and builds a list of one element.
如果对nil调用cons,Lisp将nil作为空列表对待，并构建一个含一个元素的列表。
This is very different for software, where the marginal cost for creating a copy of an existing system is almost nil.
这一点与软件有着很大的不同。
However, nil is an actual object.
然而与之不同的是，nil是作为一个对象存在的。
My foreign language skills were nil, so it had to be an English-speaking country.
我当时的外语水平完全是个零，因此能去的只能是说英语的国家。
IsUsableAction (slot) - Return 1 if an action can be used at present, nil otherwise.
如果该动作可以马上被使用，返回1，否则返回空。
IsAutoRepeatAction (slot) - Return 1 if an action is auto-repeating, nil otherwise.
如果指定动作可以自动重复，返回1，否则返回空。
IsAttackAction (slot) - Return 1 if an action is an 'attack' action (flashes during combat), nil otherwise.
如果指定的动作条是“攻击”(在战斗中闪烁)，返回1，否则返回空。
IsCurrentAction (slot) - Return 1 if an action is the one currently underway, nil otherwise.
如果指定动作正在进行中，返回1，否则返回空。
Of course, the security rendered by such an antiquated algorithm is nil.
当然，这过时的算法提供的安全是毫无价值的。
In the paper we introduce the concepts of L-trivial, nil ordered semigroups and give every ordered semigroup is L-trivial if and only if the left divisibility relation is an order.
特别地，我们还讨论了序零半群，给出了这类半群与偏序半群上的左整除关系链的关系。
Indicates if an explicit nil value can be assigned to the element.
指示是否可以为元素赋予显式零值。
Our immediate capacity for an offensive was nil.
我们尚无立刻进攻的时机。
Does it exist that there is an extremely not BCI-Subodinate to BCH-Algebra. As a result, it is the conclusion that the finite BCI-Algebra discussed in (2) is nothing but odd Nil-radical.
是否存在极不bci的BCH -代数给出了答案，同时得到一个结论：对在(2)中所讨论的有限bci -代数，均是奇诣零代数。
Does it exist that there is an extremely not BCI-Subodinate to BCH-Algebra. As a result, it is the conclusion that the finite BCI-Algebra discussed in (2) is nothing but odd Nil-radical.
是否存在极不bci的BCH -代数给出了答案，同时得到一个结论：对在(2)中所讨论的有限bci -代数，均是奇诣零代数。
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