Does a list<Tkeep track of its size or is the size re-calculated every
time you call list<T>.size()?
7  1948 
barcaroller wrote:
Does a list<Tkeep track of its size or is the size re-calculated every
time you call list<T>.size()?
The standard allows for either method. As a consequence, you should call
list<>::size() only when you have to. In particular, prefer
list<>::empty()
to
list<>::size() == 0
Best
Kai-Uwe Bux
barcaroller wrote:
Does a list<Tkeep track of its size or is the size re-calculated every
time you call list<T>.size()?
IIRC std::list::size () is *not* guaranteed to be constant-time, and in
most implementations it's indeed linear-time.
I think it has something to do with splice(), but I don't remember the
details.
On 2007-10-10 02:28, Juha Nieminen wrote:
barcaroller wrote:
>Does a list<Tkeep track of its size or is the size re-calculated every
time you call list<T>.size()?
IIRC std::list::size () is *not* guaranteed to be constant-time, and in
most implementations it's indeed linear-time.
I think it has something to do with splice(), but I don't remember the
details.
I think splice must run in O(n), where n is the number of spliced
elements, and that does not allow for recalculating the size when splicing.
--
Erik Wikström
Erik Wikström wrote:
I think splice must run in O(n), where n is the number of spliced
elements, and that does not allow for recalculating the size when splicing.
Don't you mean O(1)?
Hi
Juha Nieminen wrote:
Erik Wikström wrote:
>I think splice must run in O(n), where n is the number of spliced
elements, and that does not allow for recalculating the size when
splicing.
Don't you mean O(1)?
....where 1 is the number of spliced elements.
SCNR :-)
(It should be O(1) of course)
Markus
On 2007-10-10 15:36, Juha Nieminen wrote:
Erik Wikström wrote:
>I think splice must run in O(n), where n is the number of spliced
elements, and that does not allow for recalculating the size when splicing.
Don't you mean O(1)?
I was thinking of the version taking a position iterator, a list and two
iterators into this list. It is linear if the provided list is not
*this. Problem is I can no longer remember the reason why this prevents
size() from being O(1).
--
Erik Wikström
On 2007-10-10 05:08:10 -1000, Erik Wikström <Er***********@ telia.comsaid:
On 2007-10-10 15:36, Juha Nieminen wrote:
>Erik Wikström wrote:
>>I think splice must run in O(n), where n is the number of spliced
elements, and that does not allow for recalculating the size when splicing.
Don't you mean O(1)?
I was thinking of the version taking a position iterator, a list and two
iterators into this list. It is linear if the provided list is not
*this.
It's linear if the provided list has a different allocator from *this.
If the allocators are the same, splice does just what it's name
suggests: it munges a few pointers to snip the specified sublist out of
its current container and insert it directly into the target.
Problem is I can no longer remember the reason why this prevents
size() from being O(1).
The problem is that if splice only needs to munge the end pointers
(i.e. the allocators are the same) it doesn't know how many elements
were inserted.
--
Pete
Roundhouse Consulting, Ltd. ( www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
( www.petebecker.com/tr1book) This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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Hello all,
After perusing the Standard, I believe it is true to say that once you
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This makes a std::list<> ideal for storing vertices of an arbitrary n-ary
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