将选项与列表组合在一起会产生类型不匹配的情况，具体取决于顺序 [英] Composing Option with List in for-comprehension gives type mismatch depending on order

问题描述

为什么这种构造会在Scala中导致类型不匹配错误?

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)

如果我用列表切换Some，则可以很好地编译:

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))

这也可以正常工作:

for (first <- Some(1); second <- Some(2)) yield (first,second)

推荐答案

对于理解，此方法将转换为对 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)

成为:

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))将收到 flatMap 方法调用.由于 List 上的 flatMap 返回另一个 List ，因此for理解的结果当然将是 List .(这对我来说是新的:因为理解并不总是导致信息流，甚至不一定会导致 Seq s.)

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 Seqs.)

现在，看看在 Option 中如何声明 flatMap :

Now, take a look at how flatMap is declared in Option:

def flatMap [B] (f: (A) ⇒ Option[B]) : Option[B]

请记住这一点.让我们看看如何将理解错误(具有 Some(1)的错误)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，很容易看到 flatMap 调用的参数可以返回 List ，而不是 Option ，例如必填.

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)

这样编译就可以了.值得注意的是， Option 不是通常假定的 Seq 的子类型.

That compiles just fine. It is worth noting that Option is not a subtype of Seq, as is often assumed.

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