CODEWITHSUNDEEP
scalascala-collections
Sebastien Lorber
Sebastien Lorber

Reputation: 92170

Why is a Range transformed to a Vector after map operation?

Following Scala courses on Coursera, Martin Odersky showed an example code which is:

1 to 5 map ( i => i*i )

And he said the Range gets transformed to a Vector because they share the same interface (IndexedSeq) and the result could not be represented as a Range (it was more clear in its example since he generated a pair which is not representable as a Range).

I'm not sure to understand because I think he said previously that in a for expression the 1st generator will determine the kind of element that will be yielded, and it seems not always true, at least for Range.

And I'm not sure to understand why the output is Vector, because Vector may not be the only other one implementation that can represent the result computed above.

Can someone help me understand this part please?

Upvotes: 9

Views: 1262

Answers (3)

Rex Kerr
Rex Kerr

Reputation: 167901

map secretly takes a CanBuildFrom as an implicit argument. Its job is to produce a new collection given the one you've already got (and the type of the contents). Since Range can't contain arbitrary stuff--not even arbitrary integers--there is no CanBuildFrom that produces a Range. The most specific supertype of Range that does have a CanBuildFrom is IndexedSeq. The collection that is actually built by this is a Vector.

Upvotes: 10

Philippe
Philippe

Reputation: 9762

As I'm sure Martin also explained, for comprehensions correspond to (are translated into) chained invocations of the map and flatMap methods (and foreach if you don't use yield).

The reason why it generally results in a value of the type of the first generator is that map and flatMap generally return the same type as their receiver (map on a List returns a List, etc.).

Now the problem with Ranges is that they cannot represent things that are not regular sequences of integers. As a consequence, the return type of map and flatMap as defined for Range cannot be Range. The next best match is Vector, the prototypical implementation of an indexed sequence.

(If you look at the source code or even the Scala doc page I linked to, you will see that it is a little more complicated that just the return type, but conceptually, that is the reason. Edit: ...and now Rex Kerr just dropped the CanBuildFrom bomb.)

Upvotes: 2

adelbertc
adelbertc

Reputation: 7320

Vector is the default implementation for IndexedSeq. The map cannot be represented as a Range since the Range class is designed to contain a series of numbers that can be represented by a start, stop, and step value (similar to range in Python). The API docs specify that it's a special case of IndexSeq.

We can see 1 to 5 map { i => i * i } will get us a container of values (1, 4, 9, 16, 25). We can get a start and stop, but no constant step value.

Upvotes: 1

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