CODEWITHSUNDEEP
listperformancetcl
WilliamKF
WilliamKF

Reputation: 43129

How to efficiently get rest of a Tcl list starting from an index?

I would like to get all elements following a particular index of a list. This could be written as:

set foo {0 1 2 3 4 5 6 <...> n}
puts [lrange $foo 1 [llength $foo]]

However, it seems like a waste to compute the length of the list. It would be nice if the last argument to lrange was optional and omitting it meant to continue until the end of the list, but, alas that is not the case today.

Is there some other way of doing this efficiently in Tcl without computing the length of the list?

Upvotes: 2

Views: 541

Answers (2)

Jeff Godfrey
Jeff Godfrey

Reputation: 728

You can use "end" in place of "[llength $foo]"

So...

puts [lrange $foo 1 end]

Upvotes: 7

RHSeeger
RHSeeger

Reputation: 16262

Jeff answered your actual question well. That being said, there is one thing worth noting. Getting the length of a list (that's actually a list under the hood) is of O(1), meaning it takes no real time. The length of the list itself is stored with the metadata, and isn't recalculated. The only real cost is the overhead of a function call. Using "end" is probably still faster, just not as much as you might have thought.

But "actually a list under the hood", I mean that the interpreter is currently treating it as a list (there a deeper explanation, but not worth going into here). Since you are using [lrange] on the value, the interpreter must convert it's internal to a list... so you're pretty much guaranteed the O(1) behavior of [llength].

Upvotes: 4

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