blocks|key|2824173|text|newtype+Pair+a+b和newtype+Pair+b+a都是正确的(就像它们键入check一样)。在后一种情况下，第二种元素的类型是第一种，这有悖于直觉，尽管它在用例中可能更合适。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2824174|entityMap^0|0|G|H|G|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Both <code>newtype Pair a b</code> and <code>newtype Pair b a</code> are correct (as in, they type check). In the latter the type of the second element comes first, which is counterintuitive, although it may be more appropriate in your use case.

blocks|key|479394|text|你说的是LYAH中的定义|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|479395|newtype+Pair+b+a+=+Pair+{+getPair+::+(a,b)+}++|code-block|syntax|javascript|479396|注意前面的段落(我的强调)：|479397|事实证明，为这个实例编写实例是有点困难的。对于Maybe，我们只说实例函子，因为只有一个参数的类型构造函数才能成为函子的实例。但是，在使用(a,b)+fmap**.时，似乎没有办法对进行类似的操作，因此类型参数a最终会发生变化为了解决这个问题，我们可以通过这样一种方式来创建我们的元组，即第二个类型参数表示元组中第一个组件的类型：**|blockquote|style|CODE|BOLD|479398|因此，在本例中，我们希望(基础)对的第一个元素的类型是类型构造函数Pair中的最后一个类型。毕竟，我们希望对LYAH中的第一个参数使用fmap：|479399|--+++++++++++++++++++++b+
--+++++++++++++++++++++%7C
--+++++++++++++++++++++v++a+isn't+here…+
instance+Functor+(Pair+c)+where++
++--+but+here!
++--fmap+::+(a+->+x)+->+Pair+c+a+->+Pair+c+x++++++++++++++++++++++++
++++fmap+f+(Pair+(x,y))+=+Pair+(f+x,+y)++|479400|当然，您可以将定义更改为newtype+Pair+b+a+=+Pair+{+getPair+::+(a,b)+}。但在这个案子里会发生什么？|479401|--+++++++++++++++++++++a+
--+++++++++++++++++++++%7C
--+++++++++++++++++++++v++b+isn't+here…+
instance+Functor+(Pair+c)+where++

++--+but+here!+++++++++++++++v----+c+----v--------------------------%2B
++--fmap+::+(b+->+x)+->+Pair+c+b+->+Pair+c+x++++++++++++++++++++++++%7C
++++fmap+f+(Pair+(x,y))+=+Pair+(f+x,+y)+++++++++++++++++++--++++++++%7C
+++++++++++++++++++++++++++++--+%5E%5E%5E+uh-oh+-+that+doesn't+seem+right-%2B|479402|请记住，您的基础类型现在是(a,b)。但是，在您的Functor实例中，a现在是固定的，但是您尝试将fmap应用于此。这不管用。相反，您必须在第二个条目上使用fmap，它对应于类型构造函数Pair+a+b中的b。|479403|--+++++++++++++++++++++a+
--+++++++++++++++++++++%7C
--+++++++++++++++++++++v++b+isn't+here…+
instance+Functor+(Pair+c)+where++
++--+but+here!
++--fmap+::+(b+->+x)+->+Pair+c+b+->+Pair+c+x
++++fmap+f+(Pair+(x,y))+=+Pair+(x,+f+y)+
+++++++++++++++++--++++++++++++++++%5E%5E%5E+everything+fine+now.|479404|但此时，我们回到了最初的(a,b)实例。|479405|entityMap|0|LINK|mutability|MUTABLE|url|http://learnyouahaskell.com/functors-applicative-functors-and-monoids^0|4|8|0|0|0|0|N|5|1X|5|23|4|1X|5|23|6|2J|24|0|0|20|X|4|1V|4|0|0|0|1Y|C|18|0|0|0|2Y|D|5|P|7|10|1|1E|4|27|4|2M|8|2W|1|0|0|0|K|C|5|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@$A|1D|B|1E|1|1F]]|C|$]]|$1|D|3|E|5|F|7|1G|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1H|8|@]|9|@]|C|$]]|$1|K|3|L|5|M|7|1I|8|@$A|1J|B|1K|N|O]|$A|1L|B|1M|N|O]|$A|1N|B|1O|N|O]|$A|1P|B|1Q|N|P]|$A|1R|B|1S|N|P]|$A|1T|B|1U|N|P]]|9|@]|C|$]]|$1|Q|3|R|5|6|7|1V|8|@$A|1W|B|1X|N|P]|$A|1Y|B|1Z|N|O]|$A|20|B|21|N|O]]|9|@]|C|$]]|$1|S|3|T|5|F|7|22|8|@]|9|@]|C|$G|H]]|$1|U|3|V|5|6|7|23|8|@$A|24|B|25|N|P]|$A|26|B|27|N|O]]|9|@]|C|$]]|$1|W|3|X|5|F|7|28|8|@]|9|@]|C|$G|H]]|$1|Y|3|Z|5|6|7|29|8|@$A|2A|B|2B|N|P]|$A|2C|B|2D|N|O]|$A|2E|B|2F|N|O]|$A|2G|B|2H|N|O]|$A|2I|B|2J|N|O]|$A|2K|B|2L|N|O]|$A|2M|B|2N|N|O]|$A|2O|B|2P|N|O]]|9|@]|C|$]]|$1|10|3|11|5|F|7|2Q|8|@]|9|@]|C|$G|H]]|$1|12|3|13|5|6|7|2R|8|@$A|2S|B|2T|N|P]|$A|2U|B|2V|N|O]]|9|@]|C|$]]|$1|14|3|-4|5|6|7|2W|8|@]|9|@]|C|$]]]|15|$16|$5|17|18|19|C|$1A|1B]]]]

You're talking about the <a href="http://learnyouahaskell.com/functors-applicative-functors-and-monoids" rel="nofollow">definition in LYAH</a>:

<pre><code>newtype Pair b a = Pair { getPair :: (a,b) } 
</code></pre>

Note the paragraph that precedes it (emphasis mine):

<blockquote>
 It turns out that writing the instance for that is kind of hard. With <code>Maybe</code>, we just say instance Functor Maybe where because only type constructors that take exactly one parameter can be made an instance of Functor. But it seems like there's no way to do something like that with <code>(a,b)</code> so that the type parameter a ends up being the one that changes when we use <code>fmap</code>. To get around this, we can newtype our tuple in such a way that the second type parameter represents the type of the first component in the tuple: 
</blockquote>

So, in this case we want the type of the (underlying) pair's first element being the last type in the type constructor <code>Pair</code>. After all, we want to use <code>fmap</code> on the first argument in LYAH:

<pre><code>-- b 
-- |
-- v a isn't here… 
instance Functor (Pair c) where 
 -- but here!
 --fmap :: (a -&gt; x) -&gt; Pair c a -&gt; Pair c x 
 fmap f (Pair (x,y)) = Pair (f x, y) 
</code></pre>

You can of course change the definition to <code>newtype Pair b a = Pair { getPair :: (a,b) }</code>. But what happens in this case?

<pre><code>-- a 
-- |
-- v b isn't here… 
instance Functor (Pair c) where 

 -- but here! v---- c ----v--------------------------+
 --fmap :: (b -&gt; x) -&gt; Pair c b -&gt; Pair c x |
 fmap f (Pair (x,y)) = Pair (f x, y) -- |
 -- ^^^ uh-oh - that doesn't seem right-+
</code></pre>

Remember, your underlying type is now <code>(a,b)</code>. However, in your <code>Functor</code> instance <code>a</code> is now fixed, yet you try to apply <code>fmap</code> on that. That does not work. Instead, you have to use <code>fmap</code> on the second entry, which corresponds to <code>b</code> from the type constructor <code>Pair a b</code>:

<pre><code>-- a 
-- |
-- v b isn't here… 
instance Functor (Pair c) where 
 -- but here!
 --fmap :: (b -&gt; x) -&gt; Pair c b -&gt; Pair c x
 fmap f (Pair (x,y)) = Pair (x, f y) 
 -- ^^^ everything fine now.
</code></pre>

But at this point, we're back at the original <code>(a,b)</code> instance.

I am wondering why the following is the correct implementation of the Pair type. Specifically, why <code>Pair b a</code> and not <code>Pair a b</code>?

<pre><code>newtype Pair b a = Pair { getPair :: (a,b) } 
</code></pre>

To clarify, <code>Pair a b</code> doesn't work for the following:

<pre><code>instance Functor (Pair c) where 
fmap f (Pair (x,y)) = Pair (f x, y) 
</code></pre>

and I don't understand why.

In addition to a lot of the great answers below, I found doing the following in ghci helpful:

<pre><code>*Main&gt; newtype Pair b a = Pair (a, b) deriving (Show, Eq)
*Main&gt; :t Pair(True, "cat")
Pair(True, "cat") :: Pair [Char] Bool

*Main&gt; newtype Pair a b = Pair (a, b) deriving (Show, Eq)
*Main&gt; :t Pair(True, "cat")
Pair(True, "cat") :: Pair Bool [Char]
</code></pre>

Correct newtype definition

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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我想知道为什么下面是对类型的正确实现。具体来说，为什么Pair b a而不是Pair a bnewtype Pair b a = Pair { getPair :: (a,b) } 澄清一下，Pair a b不适用于以下方面：instance Functor (Pair c) where  fmap f (Pair (...

问正确的新型定义
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问正确的新型定义EN