blocks|key|386833|text|如果您查看任一类型，您将看到它有两个构造函数，Left和Right。为了构造Either类型的值，需要使用其中一个构造函数。例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|386834|if+x+==+1+then+Left+x|code-block|syntax|javascript|386835|具体来说，Left用于构造第一种类型(本例中为Int)，而Right用于第二种类型(在本例中为Bool)。|386836|entityMap|0|LINK|mutability|MUTABLE|url|http://hackage.haskell.org/package/base-4.12.0.0/docs/Data-Either.html^0|N|4|S|5|12|6|5|2|0|0|0|5|4|N|3|T|5|1B|4|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

If you have a look at the <a href="http://hackage.haskell.org/package/base-4.12.0.0/docs/Data-Either.html" rel="nofollow noreferrer">Either</a> type, you'll see that it has two constructors, <code>Left</code> and <code>Right</code>. In order to construct a value of the <code>Either</code> type, you need to use one of the constructors. e.g.

<pre><code>if x == 1 then Left x
</code></pre>

Specifically, <code>Left</code> is used to construct the first type of the Either type (<code>Int</code> in this case), and <code>Right</code> is used for the second type (<code>Bool</code> in this case).

blocks|key|1991651|text|给定的函数没有类型：x+::+Int和(False+:+_)+::+[Bool]，这两种类型不能统一(将::读入“有类型”)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1991652|经过调整，|1991653|findInt+(x:xs)+=
+++++if+x+==+1+then+Left+x
+++++else+Right+False+:+findInt+xs|code-block|syntax|javascript|1991654|它仍然没有类型：Left+x+::+Either+Int+b和(Right+False+:+_)+::+[Either+a+Bool]，这两种类型也不能统一。|1991655|但|1991656|findInt+::+[Int]+->+Either+Int+Bool
findInt+(x:xs)+=
+++++if+x+==+1+then+Left+x
+++++else+findInt+xs
findInt+[]+=+Right+False|1991657|具有一个类型，因为Left+x+::+Either+Int+b和Right+False+::+Either+a+Bool可以统一这两种类型：|1991658|++++++++Either+Int+b
++++++++Either+a+++Bool
+++++++-----------------++++++a+~+Int+,+b+~+Bool
++++++++Either+Int+Bool|1991659|正如类型签名所指定的那样，统一类型确实是Either+Int+Bool。|1991660|entityMap^0|A|8|J|L|1G|2|0|0|0|8|M|V|10|0|0|0|9|M|W|S|0|0|K|F|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1F|8|@]|D|@]|E|$K|L]]|$1|S|3|T|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|U|3|V|5|J|7|1L|8|@]|D|@]|E|$K|L]]|$1|W|3|X|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|Y|3|-4|5|6|7|1P|8|@]|D|@]|E|$]]]|Z|$]]

Your function as given has no type: <code>x :: Int</code> and <code>(False : _) :: [Bool]</code>, and these two types can't be unified (read <code>::</code> as "has a type").

Tweaked,

<pre><code>findInt (x:xs) =
 if x == 1 then Left x
 else Right False : findInt xs
</code></pre>

it still has no type: <code>Left x :: Either Int b</code> and <code>(Right False : _) :: [Either a Bool]</code>, and these two types can't be unified either.

But 

<pre><code>findInt :: [Int] -&gt; Either Int Bool
findInt (x:xs) =
 if x == 1 then Left x
 else findInt xs
findInt [] = Right False
</code></pre>

has a type, because <code>Left x :: Either Int b</code> and <code>Right False :: Either a Bool</code>, and these two types can be unified:

<pre><code> Either Int b
 Either a Bool
 ----------------- a ~ Int , b ~ Bool
 Either Int Bool
</code></pre>

The unified type is indeed <code>Either Int Bool</code>, as the type signature specifies.

blocks|key|1148190|text|由于您从未在输出中使用True，所以Either+Int+Bool的使用与Maybe+Int同构|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1148191|a2b+::+Either+Int+Bool+->+Maybe+Int
a2b+(Right+x)+=+Just+x
a2b+(Left+x)+=+Nothing

b2a+::+Maybe+Int+->+Either+Int+Bool
b2a+(Just+x)+->+Right+x
b2a+Nothing+->+Left+False|code-block|syntax|javascript|1148192|因此，我只需要使用Maybe+Int来简化您的函数。|1148193|findInt+::+[Int]+->+Maybe+Int
findInt+[]+=+Nothing
findInt+(x:xs)+=+if+x+==+1+then+Just+x+else+findInt+xs|1148194|entityMap^0|B|4|I|F|11|9|0|0|9|9|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

Since you never use <code>True</code> in your output, your use of <code>Either Int Bool</code> is isomorphic to <code>Maybe Int</code>:

<pre><code>a2b :: Either Int Bool -&gt; Maybe Int
a2b (Right x) = Just x
a2b (Left x) = Nothing

b2a :: Maybe Int -&gt; Either Int Bool
b2a (Just x) -&gt; Right x
b2a Nothing -&gt; Left False
</code></pre>

As such, I would just use <code>Maybe Int</code> to simplify your function.

<pre><code>findInt :: [Int] -&gt; Maybe Int
findInt [] = Nothing
findInt (x:xs) = if x == 1 then Just x else findInt xs
</code></pre>

I need to return for example either an integer or a Boolean in a recursive function.

An example code of my problem:

<pre><code>findInt :: [Int] -&gt; Either Int Bool
findInt (x:xs) =
 if x == 1 then x
 else False : findInt xs
</code></pre>

Then the error says couldn't match expected type "Either Int Bool" with actual type "Int". But I want to check whether the element is in this list and if it does it will return the element and tell me if it's not by returning a Boolean.

How to return different type in recursion function? (Haskell)

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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例如，我需要返回递归函数中的整数或布尔值。我的问题的一个示例代码：findInt :: [Int] -> Either Int BoolfindInt (x:xs) =     if x == 1 then x     else False : findInt xs然后，错误表示无法将预期的类型" Int“与实际类型”...

问如何在递归函数中返回不同的类型？(Haskell)
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