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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
InputItorForwardItadditionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector meet these requirements? And why "Legacy"?
c++
edited 2 days ago
El Profesor
11k32341
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
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
InputItorForwardItadditionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector meet these requirements? And why "Legacy"?
c++
edited 2 days ago
El Profesor
11k32341
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
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::vectorreference).
– 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
InputItorForwardItadditionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector meet these requirements? And why "Legacy"?
c++
edited 2 days ago
El Profesor
11k32341
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
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
InputItorForwardItadditionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.
Does vector meet these requirements? And why "Legacy"?
c++
c++
edited 2 days ago
El Profesor
11k32341
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
edited 2 days ago
El Profesor
11k32341
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
edited 2 days ago
El Profesor
11k32341
edited 2 days ago
El Profesor
11k32341
edited 2 days ago
El Profesor
11k32341
11k32341
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
asked Apr 9 at 7:01
Martin PerryMartin Perry
5,22333267
5,22333267
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::vectorreference).
– 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::vectorreference).
– 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).
answered Apr 9 at 7:11
bartopbartop
3,4931132
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."
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.5k53967
add a comment |
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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).
answered Apr 9 at 7:11
bartopbartop
3,4931132
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).
answered Apr 9 at 7:11
bartopbartop
3,4931132
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).
answered Apr 9 at 7:11
bartopbartop
3,4931132
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).
answered Apr 9 at 7:11
bartopbartop
3,4931132
edited Apr 9 at 7:17
answered Apr 9 at 7:11
bartopbartop
3,4931132
answered Apr 9 at 7:11
bartopbartop
3,4931132
answered Apr 9 at 7:11
bartopbartop
3,4931132
3,4931132
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."
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.5k53967
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."
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.5k53967
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."
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.5k53967
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.
34.5k53967
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.5k53967
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.5k53967
answered Apr 9 at 7:11
Nikos C.Nikos C.
34.5k53967
34.5k53967
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::vectorreference).– Some programmer dude
Apr 9 at 7:06