在 SML 中使用foldr 连接字符串
我正在尝试声明一个函数，字符串列表 ->字符串，例如输入 ["Chicago","city","USA"] 应返回“美国芝加哥市”。到目前为止我所做的是：
fun gather ts = foldr op ^ "" ts;
这似乎有点符合实际，但问题是，我想在单词之间包含空格，因为此函数将返回 "ChigagocityUSA"。
I'm trying to declare a function, string list -> string, that with the input for instance["Chicago","city","USA"] should return "Chicago city USA". What I did so far was this:
fun gather ts = foldr op ^ "" ts;
This seems to be somewhat along the lines, however the problem is, I would like to include the spaces between the words, as this function would return "ChigagocityUSA".
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是的，问题是
^是一个函数，对于两个字符串“foo”和“bar”返回“foobar”，尽管您想要“foo bar”。因此，您需要做的是定义一个函数，该函数接受两个字符串参数（作为元组）并返回两个字符串，它们之间有一个空格（因此
string1 ^ " " ^ string2）。然后，您可以将该函数作为
foldr的参数提供，并获得您想要的结果。Yes, the problem is that
^is a function that for two strings "foo" and "bar" returns "foobar", although you want "foo bar".So what you need to do is to define a function that takes two string arguments (as a tuple) and returns the two strings with a space in between them (so
string1 ^ " " ^ string2).You can then give that function as an argument to
foldrand get the result you want.使用
hd和tl获取fold的初始值。这可以避免结果中出现前导或尾随空白字符。如果您想从左到右思考，foldl很有用。定义：
用法：
Use
hdandtlto get inital values forfold. This avoids having a leading or trailing blank character in the result.foldlis useful if you want to think left-to-right.Definition:
Usage: