[英]Haskell: Treating Bool as Int
How would one implement a function that started with an int, and subtracted 1 from it for every time (going through a finite number of possibilities) one of several (for example, 5) boolean values returned 1. 如何实现一个以int开头的函数,并且每次从它中减去1(通过有限数量的可能性),其中一个(例如,5个)布尔值返回1。
How this would ideally look look is: 理想情况下看起来如何:
function list1 list2 = num
where
num = 4
- (condition from var1 = true)
- (condition from var2 = true)
- (so on, so forth as long as needed)
I have tried implementing these lines similarly to: 我尝试过类似的实现这些行:
num = startVal
- (list1conditional == desiredVal)
- (etc)
But this is returning type errors. 但这是返回类型错误。
Bool
is an instance of Enum
: you can enumerate the two values of a Bool
: False
, and then True
. Bool
是Enum
一个实例:你可以枚举Bool
的两个值: False
,然后是True
。
As a result, it implements the fromEnum :: Enum a => a -> Int
, a function that maps a value of an Enum
type to an Int
: for a Bool
, it maps False
to 0
, and True
to 1
. 其结果是,它实现了
fromEnum :: Enum a => a -> Int
,一个映射的值的函数Enum
类型的Int
:用于Bool
,它映射False
到0
,并True
至1
。
So we can use this like: 所以我们可以这样使用:
result = 5 - fromEnum cond1 - fromEnum cond2
Or for example with a list of conditions: 或者例如条件列表:
result = 5 - sum (map fromEnum [cond1, cond2, cond3])
where cond1
and cond2
, etc. are expressions of type Bool
. 其中
cond1
和cond2
等是Bool
类型的表达式。
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