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475

answers:

3

Can anyone explain why isn't the operator[] implemented for a std::list? I've searched around a bit but haven't found an answer. It wouldn't be too hard to implement or am I missing something?

+2  A: 

It would not be too hard (for the implementer) but it would be too hard at runtime, since the performance will be terrible in most cases. Forcing the user to go through each link will make it more obvious what is going on in there than 'myList[102452]' would.

florin
To elaborate a bit operator[] is a constant time operation in all the other places it is used. Giving the same name to a O(n) operation would be inconsistent and confusing.
dmckee
Well it is O(log n) in a map but I get your point.
Gab Royer
In a map it's decidedly not a positional index, which is quite obvious - except perhaps when map key is an integer, but if you're confusing positional access with key lookup, you have much bigger problems already ;)
Pavel Minaev
I know the difference, rest assured! But it remains operator[] ;).
Gab Royer
+28  A: 

Retrieving an element by index is an O(n) operation for linked list, which is what std::list is. So it was decided that providing operator[] would be deceptive, since people would be tempted to actively use it, and then you'd see code like:

 std::list<int> xs;
 for (int i = 0; i < xs.size(); ++i) {
     int x = xs[i];
     ...
 }

which is O(n^2) - very nasty. So ISO C++ standard specifically mentions that all STL sequences that support operator[] should do it in amortized constant time (23.1.1[lib.sequence.reqmts]/12), which is achievable for vector and deque, but not list.

For cases where you actually need that sort of thing, you can use std::advance algorithm:

int iter = xs.begin();
std::advance(iter, i);
int x = *iter;
Pavel Minaev
So basically, it's just a matter of preventing people of making mistakes?
Gab Royer
yep. Or of not making promises you can't keep. In the STL, the operator[] promises *efficient* access to arbitrary elements.
jalf
I see. Thanks Jalf!
Gab Royer
And thanks Pavel for the C++ standard reference!
Gab Royer
Note that std::list::size() may also be O(n) (see http://stackoverflow.com/questions/228908/is-listsize-really-on/228914), so that first loop could be O(n^2) even without the call to operator[].
bk1e
+1  A: 

I think I found the answer in another SO post http://stackoverflow.com/questions/366432/extending-stdlist

"your operator[] is O(N) time" - this is exactly why it is not included in the standard's std::list<>. – Michael Burr Dec 14 at 17:29

Still, is that the only reason?

EDIT : It seems though as people mentioned it is more a matter of consistency regarding performance then strictly performance.

Gab Royer
You mean that isn't reason enough? :-)
Chris Jester-Young
It is. Looking elsewhere, .NET `LinkedList` doesn't provide an indexer for largely the same reasons - it's just too deceptive. Traditionally, when something has a positional indexer, it is assumed that the operation is O(1).
Pavel Minaev
@Chris - No no! Rest assured! I was just wondering, because if I want to get to a certain point in a list I'd use the same O(n) algorithm, granted a list wouldn't be the most appropriate data structure then.
Gab Royer