所以我完成了在haskell中创建自己的复数数据类型。Haskell浮点错误
我也得到了,因为在这里的另一个问题,得到了一个函数,将解决一个二次方程。
现在唯一的问题是,当试图解决具有复杂根的二次方程时,代码在拥抱中生成解析错误。
即在拥抱...
Main> solve (Q 1 2 1)
(-1.0,-1.0)
Main> solve (Q 1 2 0)
(0.0,-2.0)
Main> solve (Q 1 2 2)
(
Program error: pattern match failure: v1618_v1655 (C -1.#IND -1.#IND)
它看起来像我的平方根后问题得到了应用,但我真的不知道。试图找出错误的任何帮助或任何迹象表明这个错误意味着什么将是辉煌的。
感谢,
托马斯
验证码:
-- A complex number z = (re +im.i) is represented as a pair of Floats
data Complex = C {
re :: Float,
im :: Float
} deriving Eq
-- Display complex numbers in the normal way
instance Show Complex where
show (C r i)
| i == 0 = show r
| r == 0 = show i++"i"
| r < 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r < 0 && i > 0 = show r ++ " + "++ show (C 0 i)
| r > 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r > 0 && i > 0 = show r ++ " + "++ show (C 0 i)
-- Define algebraic operations on complex numbers
instance Num Complex where
fromInteger n = C (fromInteger n) 0 -- tech reasons
(C a b) + (C x y) = C (a+x) (b+y)
(C a b) * (C x y) = C (a*x - b*y) (b*x + b*y)
negate (C a b) = C (-a) (-b)
instance Fractional Complex where
fromRational r = C (fromRational r) 0 -- tech reasons
recip (C a b) = C (a/((a^2)+(b^2))) (b/((a^2)+(b^2)))
root :: Complex -> Complex
root (C x y)
| y == 0 && x == 0 = C 0 0
| y == 0 && x > 0 = C (sqrt ((x + sqrt ((x^2) + 0))/2)) 0
| otherwise = C (sqrt ((x + sqrt ((x^2) + (y^2)))/2)) ((y/(2*(sqrt ((x + sqrt ((x^2) + (y^2)))/2)))))
-- quadratic polynomial : a.x^2 + b.x + c
data Quad = Q {
aCoeff, bCoeff, cCoeff :: Complex
} deriving Eq
instance Show Quad where
show (Q a b c) = show a ++ "x^2 + " ++ show b ++ "x + " ++ show c
solve :: Quad -> (Complex, Complex)
solve (Q a b c) = (sol (+), sol (-))
where sol op = (op (negate b) $ root $ b*b - 4*a*c)/(2 * a)
你应该让你的花车严格并且可能是双倍的,即`!Double`。 – 2012-02-16 13:48:16