【发布时间】:2021-05-13 04:18:24
【问题描述】:
所以我正在尝试学习 Elixir(我有 F# 和 Haskell 的背景),但我很难理解我的代码中发生了什么:
fizz_buzz = fn
(0, 0, _) -> "FizzBuzz"
(0, _, _) -> "Fizz"
(_, 0, _) -> "Buzz"
(_, _, c) -> c
end
fizz_buzz_rem = fn n -> fizz_buzz.(rem(n, 3), rem(n, 5), n) end
# This works
IO.puts(fizz_buzz_rem.(10))
IO.puts(fizz_buzz_rem.(11))
IO.puts(fizz_buzz_rem.(12))
IO.puts(fizz_buzz_rem.(13))
IO.puts(fizz_buzz_rem.(14))
IO.puts(fizz_buzz_rem.(15))
IO.puts(fizz_buzz_rem.(16))
IO.puts(fizz_buzz_rem.(17))
IO.puts("----------------")
inputs =
10..17
|> Enum.to_list
# Doesn't work
inputs
|> Enum.map(fizz_buzz_rem)
|> IO.puts
IO.puts("----------------")
# Doesn't work
inputs
|> Enum.map(fn n -> fizz_buzz.(rem(n, 3), rem(n, 5), n) end)
|> IO.puts
IO.puts("----------------")
manual_inputs = [10, 11, 12, 13, 14, 15, 16, 17]
# Doesn't work
manual_inputs
|> Enum.map(fizz_buzz_rem)
|> IO.puts
IO.puts("----------------")
# Doesn't work
manual_inputs
|> Enum.map(fn n -> fizz_buzz.(rem(n, 3), rem(n, 5), n) end)
|> IO.puts
IO.puts("----------------")
# The idiotic way (that doesn't work)
result = [
fizz_buzz_rem.(10),
fizz_buzz_rem.(11),
fizz_buzz_rem.(12),
fizz_buzz_rem.(13),
fizz_buzz_rem.(14),
fizz_buzz_rem.(15),
fizz_buzz_rem.(16),
fizz_buzz_rem.(17),
]
IO.puts result
# ???????????
当我运行elixir ex_02.exs 时,输出是:
Buzz
FizzBuzz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
----------------
Buzz
FizzBuzz
----------------
Buzz
FizzBuzz
----------------
Buzz
FizzBuzz
----------------
Buzz
FizzBuzz
所以你可以看到,当我将匿名函数分别应用于每个值时,我得到了正确的答案,但当我尝试使用范围、映射甚至手动将函数应用于列表的每个元素时,我最终得到了错误结果。
将匿名函数应用于 elixir 中的列表有什么问题?
【问题讨论】:
-
试试
inputs |> Enum.map(fizz_buzz_rem) |> Enum.map(&IO.puts/1)
标签: functional-programming elixir