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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 21 hours ago
Angew
135k11261354
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
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 21 hours ago
Angew
135k11261354
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
12
"Constant" meansO(1).
– Some programmer dude
21 hours ago
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
21 hours ago
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
21 hours ago
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 21 hours ago
Angew
135k11261354
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
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 21 hours ago
Angew
135k11261354
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
edited 21 hours ago
Angew
135k11261354
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
edited 21 hours ago
Angew
135k11261354
edited 21 hours ago
Angew
135k11261354
edited 21 hours ago
Angew
135k11261354
135k11261354
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
asked 22 hours ago
Martin PerryMartin Perry
5,21833267
5,21833267
12
"Constant" meansO(1).
– Some programmer dude
21 hours ago
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
21 hours ago
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
21 hours ago
add a comment |
12
"Constant" meansO(1).
– Some programmer dude
21 hours ago
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
21 hours ago
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
21 hours ago
12
12
"Constant" means
O(1).– Some programmer dude
21 hours ago
"Constant" means
O(1).– Some programmer dude
21 hours ago
1
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
21 hours ago
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
21 hours ago
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
21 hours ago
@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
21 hours ago
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 21 hours ago
bartopbartop
3,4151132
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 21 hours ago
Nikos C.Nikos C.
34.2k53967
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 21 hours ago
bartopbartop
3,4151132
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 21 hours ago
bartopbartop
3,4151132
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 21 hours ago
bartopbartop
3,4151132
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 21 hours ago
bartopbartop
3,4151132
edited 21 hours ago
answered 21 hours ago
bartopbartop
3,4151132
answered 21 hours ago
bartopbartop
3,4151132
answered 21 hours ago
bartopbartop
3,4151132
3,4151132
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 21 hours ago
Nikos C.Nikos C.
34.2k53967
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 21 hours ago
Nikos C.Nikos C.
34.2k53967
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 21 hours ago
Nikos C.Nikos C.
34.2k53967
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 21 hours ago
Nikos C.Nikos C.
34.2k53967
answered 21 hours ago
Nikos C.Nikos C.
34.2k53967
answered 21 hours ago
Nikos C.Nikos C.
34.2k53967
answered 21 hours ago
Nikos C.Nikos C.
34.2k53967
34.2k53967
add a comment |
add a comment |
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12
"Constant" means
O(1).– Some programmer dude
21 hours ago
1
I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?
– Martin Perry
21 hours ago
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
21 hours ago