Haskell将Int转换为Float [英] Haskell Converting Int to Float
问题描述
基本上我需要接受两个Int参数。
我有一个新函数的问题,它是fromIntegral函数。并返回数字的百分比,但当我运行我的代码,它不断给我这个错误:
代码:
percent :: Int - > Int - >浮动
%xy = 100 *(a`div` b)
其中a = fromIntegral x :: Float
b = fromIntegral y :: Float
错误:
没有用于(Integral Float )
使用'div'
可能的修正:为(Integral Float)
添加一个实例声明在'(*)'的第二个参数中,即`(a`div `b)'
在表达式中:100 *(a`div` b)
在'百分比'的等式中:
%xy
= 100 *(a`div `b)
其中
a = fromIntegral x :: Float
b = fromIntegral y :: Float
我读过'98 Haskell前奏,它说有这样一个叫fromInt的函数,但它从来没有工作,所以我不得不这样做,但它仍然无法正常工作。帮助!
查看 div
的类型:
div :: Integral a => a - > a - > a
您无法将您的输入转换为 Float
然后使用 div
。
使用(/)
改为:
(/)::小数a => a - > a - > a
以下代码有效:
percent :: Int - > Int - >浮动
%xy = 100 *(a / b)
其中a = fromIntegral x :: Float
b = fromIntegral y :: Float
I'm having some problem with one of the functions which I'm new at, it's the fromIntegral function.
Basically I need to take in two Int arguments and return the percentage of the numbers but when I run my code, it keeps giving me this error:
Code:
percent :: Int -> Int -> Float
percent x y = 100 * ( a `div` b )
where a = fromIntegral x :: Float
b = fromIntegral y :: Float
Error:
No instance for (Integral Float)
arising from a use of `div'
Possible fix: add an instance declaration for (Integral Float)
In the second argument of `(*)', namely `(a `div` b)'
In the expression: 100 * (a `div` b)
In an equation for `percent':
percent x y
= 100 * (a `div` b)
where
a = fromIntegral x :: Float
b = fromIntegral y :: Float
I read the '98 Haskell prelude and it says there is such a function called fromInt but it never worked so I had to go with this but it's still not working. Help!
Look at the type of div
:
div :: Integral a => a -> a -> a
You cannot transform your input to a Float
and then use div
.
Use (/)
instead:
(/) :: Fractional a => a -> a -> a
The following code works:
percent :: Int -> Int -> Float
percent x y = 100 * ( a / b )
where a = fromIntegral x :: Float
b = fromIntegral y :: Float
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