Ocaml，将所有指定的元素替换为列表中的给定元素 [英] Ocaml, replace all specified elements with a given element in a list

问题描述

我正在写一个ocaml项目，其中有一个函数用'E'替换字符列表中的所有''.这是我的建议代码:

I am writing a ocaml project, in which I have a function that replace all '' in a char-list with 'E'. Here's my code for this propose:

    let rec string_lst_change_E lst = 
    match lst with
        [] -> let a ='E'; a::[]
        |(h::t) if (h = '') -> 'E'::(string_lst_change_E t) 
        |(h::t) ->  h::(string_lst_change_E t)
;;

它说我有一个语法错误...但是我不能自己弄清楚. 我试图这样修改它:

It says I have a syntax error... But I cannot figure out by myself. I tried to modify it like this:

    let rec string_lst_change_E lst = 
    match lst with
        [] -> 'E'::[]
        |(h::t) ->if (h = '') then 'E'::(string_lst_change_E t) else h::(string_lst_change_E t)
;;

但是仍然存在语法错误...(在线|(h :: t)-> .... char 18-21)

but still there's syntax error...(on the line |(h::t) -> .... char 18-21)

请帮助我看看.谢谢！

Please help me to take a look at it. Thank you!

推荐答案

这是第一个错误所在:[] -> let a ='E'; a::[]如果要在声明后使用a，则应改写[] -> let a = 'E' in a ::[].显然，[] -> ['E']更简单.

This is where the first error lies: [] -> let a ='E'; a::[] If you want to use a after declaring it, you should instead write [] -> let a = 'E' in a ::[]. Obviously, [] -> ['E'] is simpler.

第二个是在模式匹配中使用if.您应该改用when:|(h::t) when h = '' -> 'E'::(string_lst_change_E t)

The second is the use of if in a pattern match. You should use when instead: |(h::t) when h = '' -> 'E'::(string_lst_change_E t)

但是什么是?"空字符?你怎么会得到一个字符串?键入''本身是语法错误.尝试在顶级！为了使您的代码编译，我用' '替换了''.

But what's '' anyway? The empty character? How would you get this in a string? Typing '' is itself a syntax error. Try it in the toplevel! To make your code compile, I replaced '' by ' '.

let rec string_lst_change_E lst =
    match lst with
        | [] -> let a ='E' in a::[]
        | (h::t) when h = ' ' -> 'E'::(string_lst_change_E t) 
        | (h::t) ->  h::(string_lst_change_E t)

请注意，您可以在此处简单地使用功能

Note that you can simply use function here:

let rec string_lst_change_E = function
    | [] -> let a ='E' in a::[]
    | (h::t) when h = ' ' -> 'E'::(string_lst_change_E t) 
    | (h::t) ->  h::(string_lst_change_E t)

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