haskell中负整数的除法规则是什么？

Question

在 GHCI 中，我得到了这些结果：

div (-7) ( 2) = -4
div ( 7) (-2) = -4
div (-7) (-2) =  3

mod ( 7)  ( 2) =  1
mod ( 7)  (-2) = -1
mod (-7)  ( 2) =  1
mod (-7)  (-2) = -1

和div 4 (-3)是-2，mod 4 (-3)是-2。

我不明白如何得到这些结果。有一些法律或规则吗？正确的计算步骤是什么？

Answer 1

通常最好先查看文档。对于mod函数，我们看到：

mod :: a -> a -> a

integer modulus, satisfying
(x `div` y)*y + (x `mod` y) == x

所以x `mod` y计算为：

mod x y = x - y * (div x y) -- logically equivalent definition

并且：

div :: a -> a -> a

integer division truncated toward negative infinity.

所以如果除法是 -3.5，div 将 return -4。

所以：

mod 7 2       = 7 - 2*(div 7 2)         = 7 - 2 * 3       = 7 - 6    = 1
mod 7 (-2)    = 7 - (-2)*(div 7 (-2))   = 7 - (-2) * (-4) = 7 - 8    = -1
mod (-7) 2    = (-7) - 2 * (div (-7) 2) = (-7) - 2 * (-4) = (-7) + 8 = 1
mod (-7) (-2) = (-7) - (-2) * (div (-7) (-2)) = -7 - (-2) * 3 = -7 + 6 = -1

haskell中负整数的除法规则是什么？

What's the division rules of negative int numbers in haskell?

haskell

division

modulo