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Is std::next for vector O(n) or O(1)?
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In C++11 I use std::next
because If I want to change vector
to list
, I dont have to change the rest of code.
For list
, std::next
is O(n), because I need to iterate all elements. But how is it for a vector
? I have found this note on cppreference:
However, if
InputIt
orForwardIt
additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector
meet these requirements? And why "Legacy"?
c++
add a comment |
In C++11 I use std::next
because If I want to change vector
to list
, I dont have to change the rest of code.
For list
, std::next
is O(n), because I need to iterate all elements. But how is it for a vector
? I have found this note on cppreference:
However, if
InputIt
orForwardIt
additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector
meet these requirements? And why "Legacy"?
c++
13
"Constant" meansO(1)
.
– Some programmer dude
Apr 9 at 7:03
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
Apr 9 at 7:04
4
@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in thestd::vector
reference).
– Some programmer dude
Apr 9 at 7:06
add a comment |
In C++11 I use std::next
because If I want to change vector
to list
, I dont have to change the rest of code.
For list
, std::next
is O(n), because I need to iterate all elements. But how is it for a vector
? I have found this note on cppreference:
However, if
InputIt
orForwardIt
additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector
meet these requirements? And why "Legacy"?
c++
In C++11 I use std::next
because If I want to change vector
to list
, I dont have to change the rest of code.
For list
, std::next
is O(n), because I need to iterate all elements. But how is it for a vector
? I have found this note on cppreference:
However, if
InputIt
orForwardIt
additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector
meet these requirements? And why "Legacy"?
c++
c++
edited Apr 11 at 11:26
El Profesor
11k32341
11k32341
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,31633269
5,31633269
13
"Constant" meansO(1)
.
– Some programmer dude
Apr 9 at 7:03
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
Apr 9 at 7:04
4
@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in thestd::vector
reference).
– Some programmer dude
Apr 9 at 7:06
add a comment |
13
"Constant" meansO(1)
.
– Some programmer dude
Apr 9 at 7:03
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
Apr 9 at 7:04
4
@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in thestd::vector
reference).
– Some programmer dude
Apr 9 at 7:06
13
13
"Constant" means
O(1)
.– Some programmer dude
Apr 9 at 7:03
"Constant" means
O(1)
.– Some programmer dude
Apr 9 at 7:03
1
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
Apr 9 at 7:04
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
Apr 9 at 7:04
4
4
@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the
std::vector
reference).– Some programmer dude
Apr 9 at 7:06
@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the
std::vector
reference).– Some programmer dude
Apr 9 at 7:06
add a comment |
2 Answers
2
active
oldest
votes
There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator
or RandomAccessIterator
. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator
and so on for pre-concept requirements and RandomAccessIterator
and so for concept requirements.
And so yes, std::vector::iterator
meets requirements of LegacyRandomAccessIterator
and actually will be fulfilling RandomAccessIterator
concept as well. This leads straight to conclusion that std::next
called on vector::iterator
has complexity O(1).
add a comment |
Does vector meet these requirements?
Yes, it does:
https://en.cppreference.com/w/cpp/container/vector
Quote: "iterator LegacyRandomAccessIterator"
And why "Legacy"?
The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator
or RandomAccessIterator
. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator
and so on for pre-concept requirements and RandomAccessIterator
and so for concept requirements.
And so yes, std::vector::iterator
meets requirements of LegacyRandomAccessIterator
and actually will be fulfilling RandomAccessIterator
concept as well. This leads straight to conclusion that std::next
called on vector::iterator
has complexity O(1).
add a comment |
There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator
or RandomAccessIterator
. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator
and so on for pre-concept requirements and RandomAccessIterator
and so for concept requirements.
And so yes, std::vector::iterator
meets requirements of LegacyRandomAccessIterator
and actually will be fulfilling RandomAccessIterator
concept as well. This leads straight to conclusion that std::next
called on vector::iterator
has complexity O(1).
add a comment |
There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator
or RandomAccessIterator
. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator
and so on for pre-concept requirements and RandomAccessIterator
and so for concept requirements.
And so yes, std::vector::iterator
meets requirements of LegacyRandomAccessIterator
and actually will be fulfilling RandomAccessIterator
concept as well. This leads straight to conclusion that std::next
called on vector::iterator
has complexity O(1).
There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator
or RandomAccessIterator
. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator
and so on for pre-concept requirements and RandomAccessIterator
and so for concept requirements.
And so yes, std::vector::iterator
meets requirements of LegacyRandomAccessIterator
and actually will be fulfilling RandomAccessIterator
concept as well. This leads straight to conclusion that std::next
called on vector::iterator
has complexity O(1).
edited Apr 9 at 7:17
answered Apr 9 at 7:11
bartopbartop
3,5131132
3,5131132
add a comment |
add a comment |
Does vector meet these requirements?
Yes, it does:
https://en.cppreference.com/w/cpp/container/vector
Quote: "iterator LegacyRandomAccessIterator"
And why "Legacy"?
The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."
add a comment |
Does vector meet these requirements?
Yes, it does:
https://en.cppreference.com/w/cpp/container/vector
Quote: "iterator LegacyRandomAccessIterator"
And why "Legacy"?
The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."
add a comment |
Does vector meet these requirements?
Yes, it does:
https://en.cppreference.com/w/cpp/container/vector
Quote: "iterator LegacyRandomAccessIterator"
And why "Legacy"?
The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."
Does vector meet these requirements?
Yes, it does:
https://en.cppreference.com/w/cpp/container/vector
Quote: "iterator LegacyRandomAccessIterator"
And why "Legacy"?
The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.8k53967
34.8k53967
add a comment |
add a comment |
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13
"Constant" means
O(1)
.– Some programmer dude
Apr 9 at 7:03
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
Apr 9 at 7:04
4
@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the
std::vector
reference).– Some programmer dude
Apr 9 at 7:06