我想要类似的东西
splitBy pred list = ( filter pred list, filter (not . pred) list )
但一次通过.
假设我有两个功能,f:X->Y并且g:Y*Y->Z.我想做第三个功能,h(a, b) = g(f(a), f(b)).
h a b = g (f a) (f b)
有什么方法可以写出来h(a, b) = g*f (a, b)吗？
如果h(a,b,c,d) = g2*g1*f2*f1 (a,b,c,d),在哪里g_i需要2个参数？
有人可以解释第二个结果吗？
user$ set 5 5
user$ n=2
user$ eval echo \$$n
5
user$ echo `eval echo \$$n`
10268n
10268是bash pid.
GNU bash,版本4.0.35(0)-release(i386-portbld-freebsd7.2)
UPD:这很好用:
user$ echo `eval echo \\$$n`
5
但是之后...
user$ echo `eval echo \\\$$n` #3
5
user$ echo `eval echo \\\\$$n` #4
10268n
user$ echo `eval echo \\\\\$$n` #5
10268n
user$ echo `eval echo \\\\\\$$n` #6
$2
user$ echo `eval echo \\\\\\\$$n` #7
$2
user$ echo `eval echo \\\\\\\\$$n` #8
$2
user$ echo `eval echo \\\\\\\\\$$n` #9
10268n
我想弄清楚什么是错的.似乎类型有问题,但单独使用相同的表达式可以正常工作.
这是代码:
a = [9, 4, 12, 0, -6, 16] :: [Int]
qsort:: [Int] -> [Int]
qsort [] = []
qsort [x] = [x]
qsort xs = (qsort l)++(qsort r)
where m = (realToFrac(sum xs)) / (realToFrac(length xs))
l = filter (<=m) xs
r = filter (>m) xs
main::IO()
main = do
print (show (qsort a))
它抛出:
main.hs:7:36:
No instance for (Fractional Int)
arising from a use of `/'
Possible fix: add an instance declaration for (Fractional Int)
In the …