在 for-comprehension 中使用 List 组合选项会根据顺序给出类型不匹配 [英] Composing Option with List in for-comprehension gives type mismatch depending on order
问题描述
为什么这种构造会导致 Scala 中出现类型不匹配错误?
Why does this construction cause a Type Mismatch error in Scala?
for (first <- Some(1); second <- List(1,2,3)) yield (first,second)
<console>:6: error: type mismatch;
found : List[(Int, Int)]
required: Option[?]
for (first <- Some(1); second <- List(1,2,3)) yield (first,second)
如果我用 List 切换 Some ,它编译得很好:
If I switch the Some with the List it compiles fine:
for (first <- List(1,2,3); second <- Some(1)) yield (first,second)
res41: List[(Int, Int)] = List((1,1), (2,1), (3,1))
这也很好用:
for (first <- Some(1); second <- Some(2)) yield (first,second)
推荐答案
For comprehensions 转换为对 map
或 flatMap
方法的调用.例如这个:
For comprehensions are converted into calls to the map
or flatMap
method. For example this one:
for(x <- List(1) ; y <- List(1,2,3)) yield (x,y)
变成:
List(1).flatMap(x => List(1,2,3).map(y => (x,y)))
因此,第一个循环值(在本例中为 List(1)
)将接收 flatMap
方法调用.由于List
上的flatMap
返回另一个List
,for comprehension 的结果当然是一个List
.(这对我来说是新的:因为理解并不总是导致流,甚至不一定在 Seq
中.)
Therefore, the first loop value (in this case, List(1)
) will receive the flatMap
method call. Since flatMap
on a List
returns another List
, the result of the for comprehension will of course be a List
. (This was new to me: For comprehensions don't always result in streams, not even necessarily in Seq
s.)
现在,看看flatMap
是如何在Option
中声明的:
Now, take a look at how flatMap
is declared in Option
:
def flatMap [B] (f: (A) ⇒ Option[B]) : Option[B]
记住这一点.让我们看看理解的错误(带有 Some(1)
的错误)如何转换为一系列 map 调用:
Keep this in mind. Let's see how the erroneous for comprehension (the one with Some(1)
) gets converted to a sequence of map calls:
Some(1).flatMap(x => List(1,2,3).map(y => (x, y)))
现在,很容易看出 flatMap
调用的参数返回的是 List
,而不是 Option
,如需要.
Now, it's easy to see that the parameter of the flatMap
call is something that returns a List
, but not an Option
, as required.
为了解决这个问题,您可以执行以下操作:
In order to fix the thing, you can do the following:
for(x <- Some(1).toSeq ; y <- List(1,2,3)) yield (x, y)
编译就好了.值得注意的是,Option
不是 Seq
的子类型,正如人们通常认为的那样.
That compiles just fine. It is worth noting that Option
is not a subtype of Seq
, as is often assumed.
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