Haskell中的字符串插值

Question

我想编写将输出的代码:

length [1,2,3] => 3   

在Ruby中,我可以这样做:

puts "length [1,2,3] => #{[1,2,3].length}"

以下尝试是Haskell失败了......

Prelude Data.List> print "length [1,2,3]"
"length [1,2,3]"
Prelude Data.List> print (length [1,2,3])
3
Prelude Data.List> print "length [1,2,3]" (length [1,2,3])

<interactive>:1:0:
  Couldn't match expected type `Int -> t'
     against inferred type `IO ()'
 In the expression: print "length [1,2,3]" (length [1, 2, 3])
 In the definition of `it':
      it = print "length [1,2,3]" (length [1, 2, 3])
 Prelude Data.List>

Answer 1

您也可以使用Text.PrintfGHC基础库中包含的内容:

> let format s = printf "length %s => %d\n" (show s) (length s)
> format [1,2,3]
length [1,2,3] => 3

如果你想要更好的情况,Hackage http://hackage.haskell.org上有几个字符串插值包.

Answer 2

虽然这里的其他海报提到了许多"正确"的方法来进行字符串插值,但使用quasiquotation和interpolatedstring-perl6库有一种更好的方式:

{-# LANGUAGE QuasiQuotes, ExtendedDefaultRules #-}

import Text.InterpolatedString.Perl6 (qq)

main = putStrLn [$qq| length [1,2,3] => ${length [1,2,3]} |]

实际上还有一个interpolatedstring-qq库,它提供了Ruby语法.

{-# LANGUAGE QuasiQuotes, ExtendedDefaultRules #-}

import Text.InterpolatedString.QQ (istr)

main = putStrLn [$istr| length [1,2,3] => #{length [1,2,3]} |]

也就是说,您可能应该使用show和++或concat将字符串粘合在一起

main = putStrLn $ "length [1,2,3] => " ++ show (length [1,2,3])

要么

main = putStrLn $ concat ["length [1,2,3] => ", show $ length (1,2,3)]

当你将许多字符串片段粘在一起时,后者往往看起来更好,代码方面.

Answer 3

字符串实际上只是列表.因此,您可以转换从长度返回的数字,并将其附加到具有正常列表函数的其他字符串:

print $ "length [1,2,3] " ++ show (length [1,2,3])

Answer 4

使用格式从功能的文本格式,简单的库:

import Text.Format
format "length [1,2,3] => {0}" [show $ length [1,2,3]]