haskell之乘以零时的惰性求值

通过下面的程序，

f 0 0 0 1 = 0 
f 0 0 1 0 = f 0 0 0 1 + 1 
f 0 1 0 0 = f 0 0 1 1 + 1 
f 1 0 0 0 = f 0 1 1 1 + 1 
f a b c d = (p + q + r + s) / (a + b + c + d) 
    where 
    p 
        | a > 0 = a * f (a - 1) (b + 1) (c + 1) (d + 1) 
        | otherwise = 0 
    q 
        | b > 0 = b * f a (b - 1) (c + 1) (d + 1) 
        | otherwise = 0 
    r 
        | c > 0 = c * f a b (c - 1) (d + 1) 
        | otherwise = 0 
    s 
        | d > 0 = d * f a b c (d - 1) 
        | otherwise = 0 
 
main = print (f 1 1 1 1) 

f 0 0 0 1 = 0 
f 0 0 1 0 = f 0 0 0 1 + 1 
f 0 1 0 0 = f 0 0 1 1 + 1 
f 1 0 0 0 = f 0 1 1 1 + 1 
f a b c d = (p + q + r + s) / (a + b + c + d) 
    where 
    p = a * f (a - 1) (b + 1) (c + 1) (d + 1) 
    q = b * f a (b - 1) (c + 1) (d + 1) 
    r = c * f a b (c - 1) (d + 1) 
    s = d * f a b c (d - 1) 
 
main = print (f 1 1 1 1) 

请您参考如下方法:

内置乘法对两个参数都是严格的——也就是说，它计算两个参数——不管它们中的一个是否为零，这就是导致程序循环的原因。您可以定义自己的乘法运算符，它可以懒惰地消除一个或另一个参数:

0 .* y = 0 
x .* y = x * y 

x .* y = unambs [ assuming (x == 0) 0 
                , assuming (y == 0) 0 
                , x * y 
                ]