问题描述
我无法进行此操作。我正在使用override init() {
super.init()
let center = UNUserNotificationCenter.current()
center.getNotificationSettings { (settings) in
if(settings.authorizationStatus == .authorized)
{
self.allowednotifications = true
self.showResult = true
}
else if (settings.authorizationStatus == .denied)
{
self.allowednotifications = false
self.showResult = true
}
}
}
方法定义谓词,该方法评估列表x和列表y的元素是否属于相同类型且位于相同位置。当我尝试调用谓词进行测试时,问题就来了。我收到错误消息deftype SameType(x y)
,这是我的代码:
ERROR: SameType is undefined
)
这就是我的称呼方式
(deftype SameType (x y)
`(cond
((and (null x) (null y) T))
(
(and (numberp (car x)) (numberp (car y)))
(SameType (cdr x) (cdr y) )
)
(
(and (stringp (car x)) (stringp (car y)))
(SameType (cdr x) (cdr y) )
)
(
(and (atom (car x)) (atom (car y)))
(SameType (cdr x) (cdr y) )
)
(T nil)
)
我已经检查了各种onine资源,甚至在此站点上也有相关问题。
解决方法
deftype
可用于定义类型,而不是谓词。例如,要仅使用整数定义列表的类型,可以编写如下内容:
(defun intlistp (l)
"predicate to check if l is a list consisting only of integers"
(and (listp l) ; l is a list and
(every #'integerp l))) ; every element of l is an integer
(deftype integer-list ()
"the type of list of integers"
`(satisfies intlistp))
,然后您可以检查一个值是否满足此类型:
CL-USER> (typep '(1 2 3) 'integer-list)
T
CL-USER> (typep '(1 2.5 3) 'integer-list)
NIL
如果要根据定义检查两个列表的类型是否相同,则可以定义一个常规函数:
(defun same-type (l1 l2)
"check if lists l1 and l2 have the same length and corresponding
elements of the same CL type"
(cond ((null l1) ; if l1 is null
(null l2)) ; returns true only if also l2 is null
((and (consp l1) ; if l1 is a cons
(consp l2) ; and l2 is a cons too,(typep (car l1) (type-of (car l2)))) ; and their cars have the same CL type
(same-type (cdr l1) (cdr l2))))) ; go recursively on their cdrs
CL-USER> (same-type '(1 a 3) '(2 b 4))
T
CL-USER> (same-type '(1 "a" 3) '(2 "b" 3))
T
CL-USER> (same-type '(1 a 3) '(2 b 4.5))
NIL
CL-USER> (same-type '(1 a 3) '(2 b 4 3))
NIL
CL-USER> (same-type '(1 2 (3 4)) '(1 6 (4 5)))
T
CL-USER> (same-type '(1 2 (3 4)) '(1 6 (4 5 6)))
T
请注意,从上一个示例中可以看到,仅针对列表的第一级检查类型。