Because seq is defined as returning nil for an empty sequence. The
only way to find that out for a lazy sequence is to realize the first
element.
I'm not sure if that answers why seq should realize the first
element. Even by what you say, only if I wanted to find if my LazySeq
was nil should I
@Richard
>> 1. In case coll is a LazySeq why does (seq coll) realize its first
>> element? I thought seq just did a type conversion and all of list,
>> vector .. etc implemented Seqable or something.
> Because seq is defined as returning nil for an empty sequence. The
> only way to find that out
On May 30, 8:11 pm, Richard Newman wrote:
> > 2. Why is there no other way to determine an empty coll except (not
> > (seq coll)).
>
> user=> (empty? [])
> true
And in fact, the docs for (empty?) say:
"Please use the idiom (seq x) rather than (not (empty? x))"
Perhaps the seq docs should indi
1. In case coll is a LazySeq why does (seq coll) realize its first
element? I thought seq just did a type conversion and all of list,
vector .. etc implemented Seqable or something.
Because seq is defined as returning nil for an empty sequence. The
only way to find that out for a lazy sequenc
Hi Clojurians,
I have some conceptual questions on the sequence abstraction. I
understand that (seq coll) will give me a "sequence".
coll maybe be a list, vector, map, set, LazySeq or nil.
1. In case coll is a LazySeq why does (seq coll) realize its first
element? I thought seq just did a type
