我需要用矩阵乘法求解一个线性方程。我正在使用拥抱，所以我尝试了以下操作：
import Matrix(multiplyMat)
// more code follows
当我在控制台中调用此命令时，控制台会告诉我：
找不到导入的模块"备注"。
但是它在所有模块的目录中，那么为什么会出现错误呢？
该文件名为Matrix.hs:
-- Some simple Hugs programs for manipulating matrices.
--
module Matrix where
import List
type Matrix k = [Row k] -- matrix represented by a list of its rows
type Row k = [k] -- a row represented by a list of literals
-- General utility functions:
shapeMat :: Matrix k -> (Int, Int)
shapeMat mat = (rows mat, cols mat)
rows :: Matrix k -> Int
rows mat = length mat
cols :: Matrix k -> Int
cols mat = length (head mat)
idMat :: Int -> Matrix Int
idMat 0 = []
idMat (n+1) = [1:replicate n 0] ++ map (0:) (idMat n)
-- Matrix multiplication:
multiplyMat :: Matrix Int -> Matrix Int -> Matrix Int
multiplyMat a b | cols a==rows b = [[row `dot` col | col<-b'] | row<-a]
| otherwise = error "incompatible matrices"
where v `dot` w = sum (zipWith (*) v w)
b' = transpose b
-- An attempt to implement the standard algorithm for converting a matrix
-- to echelon form...
echelon :: Matrix Int -> Matrix Int
echelon rs
| null rs || null (head rs) = rs
| null rs2 = map (0:) (echelon (map tail rs))
| otherwise = piv : map (0:) (echelon rs')
where rs' = map (adjust piv) (rs1++rs3)
(rs1,rs2) = span leadZero rs
leadZero (n:_) = n==0
(piv:rs3) = rs2
-- To find the echelon form of a matrix represented by a list of rows rs:
--
-- {first line in definition of echelon}:
-- If either the number of rows or the number of columns in the matrix
-- is zero (i.e. if null rs || null (head rs)), then the matrix is
-- already in echelon form.
--
-- {definition of rs1, rs2, leadZero in where clause}:
-- Otherwise, split the matrix into two submatrices rs1 and rs2 such that
-- rs1 ++ rs2 == rs and all of the rows in rs1 begin with a zero.
--
-- {second line in definition of echelon}:
-- If rs2 is empty (i.e. if null rs2) then every row begins with a zero
-- and the echelon form of rs can be found by adding a zero on to the
-- front of each row in the echelon form of (map tail rs).
--
-- {Third line in definition of echelon, and definition of piv, rs3}:
-- Otherwise, the first row of rs2 (denoted piv) contains a non-zero
-- leading coefficient. After moving this row to the top of the matrix
-- the original matrix becomes piv:(rs1++rs3).
-- By subtracting suitable multiples of piv from (suitable multiples of)
-- each row in (rs1++rs3) {see definition of adjust below}, we obtain a
-- matrix of the form:
--
-- <----- piv ------>
-- __________________
-- 0 |
-- . |
-- . | rs' where rs' = map (adjust piv) (rs1++rs3)
-- . |
-- 0 |
--
-- whose echelon form is piv : map (0:) (echelon rs').
--
adjust :: Num a => Row a -> Row a -> Row a
adjust (m:ms) (n:ns) = zipWith (-) (map (n*) ms) (map (m*) ns)
-- A more specialised version of this, for matrices of integers, uses the
-- greatest common divisor function gcd in an attempt to try and avoid
-- result matrices with very large coefficients:
--
-- (I'm not sure this is really worth the trouble!)
adjust' :: Row Int -> Row Int -> Row Int
adjust' (m:ms) (n:ns) = if g==0 then ns
else zipWith (x y -> b*y - a*x) ms ns
where g = gcd m n
a = n `div` g
b = m `div` g
-- end!!
您的Matrix文件不导出任何内容。
要导出，请这样写标题：
module Module (function, value, Type(Constructor))