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Why/when should I use std::unique/shared_ptr (std::vector) over just std::vector?


What is a smart pointer and when should I use one?Concatenating two std::vectorsWhen should static_cast, dynamic_cast, const_cast and reinterpret_cast be used?How to find out if an item is present in a std::vector?Why is “using namespace std” considered bad practice?What is std::move(), and when should it be used?Why should C++ programmers minimize use of 'new'?Are the days of passing const std::string & as a parameter over?Why is my program slow when looping over exactly 8192 elements?Why should I use a pointer rather than the object itself?













-1















I'm a little bit confused about the main use of std::unique/shared_ptr(std::vector<>) when I can simply use a std::vector<>, which, as I know, is itself inherently a dynamic array. As I have also seen around, people say that there is no any performance difference between these two. So, based on all this, what is the point of using a smart pointer pointing to a container (in this case, a vector) instead of a vector alone?










share|improve this question



















  • 5





    Do you need to shared ownership of the vector ?

    – Jarod42
    Mar 7 at 21:34






  • 1





    people say that there is no any performance difference between these two Whoever told you that lied, or didn't know what they were talking about. std::shared_ptr is expensive and should only be used is you really need shared ownership. What is your use case for your vector?

    – NathanOliver
    Mar 7 at 21:36











  • Which context? What is 'TYPE'?

    – SergeyA
    Mar 7 at 21:40











  • whereas std::shared_ptr<std::vector<T>> might make sense, I don't see the point of std::unique_ptr<std::vector<T>>.

    – Jarod42
    Mar 7 at 21:40











  • Thank you. No, my problem is not sharing the object (that's why I just added the unique). I just wanted to know when we can have a dynamically allocated vector/array, why should we use smart pointers whose, as I know, main application is dynamic allocation?

    – Federico
    Mar 7 at 21:41















-1















I'm a little bit confused about the main use of std::unique/shared_ptr(std::vector<>) when I can simply use a std::vector<>, which, as I know, is itself inherently a dynamic array. As I have also seen around, people say that there is no any performance difference between these two. So, based on all this, what is the point of using a smart pointer pointing to a container (in this case, a vector) instead of a vector alone?










share|improve this question



















  • 5





    Do you need to shared ownership of the vector ?

    – Jarod42
    Mar 7 at 21:34






  • 1





    people say that there is no any performance difference between these two Whoever told you that lied, or didn't know what they were talking about. std::shared_ptr is expensive and should only be used is you really need shared ownership. What is your use case for your vector?

    – NathanOliver
    Mar 7 at 21:36











  • Which context? What is 'TYPE'?

    – SergeyA
    Mar 7 at 21:40











  • whereas std::shared_ptr<std::vector<T>> might make sense, I don't see the point of std::unique_ptr<std::vector<T>>.

    – Jarod42
    Mar 7 at 21:40











  • Thank you. No, my problem is not sharing the object (that's why I just added the unique). I just wanted to know when we can have a dynamically allocated vector/array, why should we use smart pointers whose, as I know, main application is dynamic allocation?

    – Federico
    Mar 7 at 21:41













-1












-1








-1








I'm a little bit confused about the main use of std::unique/shared_ptr(std::vector<>) when I can simply use a std::vector<>, which, as I know, is itself inherently a dynamic array. As I have also seen around, people say that there is no any performance difference between these two. So, based on all this, what is the point of using a smart pointer pointing to a container (in this case, a vector) instead of a vector alone?










share|improve this question
















I'm a little bit confused about the main use of std::unique/shared_ptr(std::vector<>) when I can simply use a std::vector<>, which, as I know, is itself inherently a dynamic array. As I have also seen around, people say that there is no any performance difference between these two. So, based on all this, what is the point of using a smart pointer pointing to a container (in this case, a vector) instead of a vector alone?







c++ c++11






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 7 at 21:42







Federico

















asked Mar 7 at 21:33









FedericoFederico

47110




47110







  • 5





    Do you need to shared ownership of the vector ?

    – Jarod42
    Mar 7 at 21:34






  • 1





    people say that there is no any performance difference between these two Whoever told you that lied, or didn't know what they were talking about. std::shared_ptr is expensive and should only be used is you really need shared ownership. What is your use case for your vector?

    – NathanOliver
    Mar 7 at 21:36











  • Which context? What is 'TYPE'?

    – SergeyA
    Mar 7 at 21:40











  • whereas std::shared_ptr<std::vector<T>> might make sense, I don't see the point of std::unique_ptr<std::vector<T>>.

    – Jarod42
    Mar 7 at 21:40











  • Thank you. No, my problem is not sharing the object (that's why I just added the unique). I just wanted to know when we can have a dynamically allocated vector/array, why should we use smart pointers whose, as I know, main application is dynamic allocation?

    – Federico
    Mar 7 at 21:41












  • 5





    Do you need to shared ownership of the vector ?

    – Jarod42
    Mar 7 at 21:34






  • 1





    people say that there is no any performance difference between these two Whoever told you that lied, or didn't know what they were talking about. std::shared_ptr is expensive and should only be used is you really need shared ownership. What is your use case for your vector?

    – NathanOliver
    Mar 7 at 21:36











  • Which context? What is 'TYPE'?

    – SergeyA
    Mar 7 at 21:40











  • whereas std::shared_ptr<std::vector<T>> might make sense, I don't see the point of std::unique_ptr<std::vector<T>>.

    – Jarod42
    Mar 7 at 21:40











  • Thank you. No, my problem is not sharing the object (that's why I just added the unique). I just wanted to know when we can have a dynamically allocated vector/array, why should we use smart pointers whose, as I know, main application is dynamic allocation?

    – Federico
    Mar 7 at 21:41







5




5





Do you need to shared ownership of the vector ?

– Jarod42
Mar 7 at 21:34





Do you need to shared ownership of the vector ?

– Jarod42
Mar 7 at 21:34




1




1





people say that there is no any performance difference between these two Whoever told you that lied, or didn't know what they were talking about. std::shared_ptr is expensive and should only be used is you really need shared ownership. What is your use case for your vector?

– NathanOliver
Mar 7 at 21:36





people say that there is no any performance difference between these two Whoever told you that lied, or didn't know what they were talking about. std::shared_ptr is expensive and should only be used is you really need shared ownership. What is your use case for your vector?

– NathanOliver
Mar 7 at 21:36













Which context? What is 'TYPE'?

– SergeyA
Mar 7 at 21:40





Which context? What is 'TYPE'?

– SergeyA
Mar 7 at 21:40













whereas std::shared_ptr<std::vector<T>> might make sense, I don't see the point of std::unique_ptr<std::vector<T>>.

– Jarod42
Mar 7 at 21:40





whereas std::shared_ptr<std::vector<T>> might make sense, I don't see the point of std::unique_ptr<std::vector<T>>.

– Jarod42
Mar 7 at 21:40













Thank you. No, my problem is not sharing the object (that's why I just added the unique). I just wanted to know when we can have a dynamically allocated vector/array, why should we use smart pointers whose, as I know, main application is dynamic allocation?

– Federico
Mar 7 at 21:41





Thank you. No, my problem is not sharing the object (that's why I just added the unique). I just wanted to know when we can have a dynamically allocated vector/array, why should we use smart pointers whose, as I know, main application is dynamic allocation?

– Federico
Mar 7 at 21:41












1 Answer
1






active

oldest

votes


















2














First of all, you shouldn't be using std::shared_ptr unless you need the specific "shared ownership" semantics associated with std::shared_ptr. If you need a smart pointer, you should default to std::unique_ptr by default, and only switch away from it in the scenario where you expressly find that you need to.



Secondly: ostensibly, the reason to prefer std::unique_ptr<TYPE> over TYPE is if you plan to move the object around a lot. This is the common design paradigm for large objects that are either unmovable, or otherwise expensive to move—i.e. they implemented a Copy Constructor and didn't implement a Move Constructor, so moves are forced to behave like a Copy.



std::vector, however, does have relatively efficient move semantics: if you move a std::vector around, regardless of how complex its contained types are, the move only constitutes a couple of pointer swaps. There's no real risk that moving a std::vector will incur a large amount of computational complexity. Even in the scenario where you're overriding a previously allocated array (invoking the Destructors of all objects in the vector), you'd still have that complexity if you were using std::unique_ptr<std::vector<TYPE>> instead, saving you nothing.



There are two advantages to std::unique_ptr<std::vector<TYPE>>. The first of which is that it gets rid of the implicit copy constructor; maybe you want to enforce to maintaining programmers that the object shouldn't be copied. But that's a pretty niche use. The other advantage is that it allows you to stipulate the scenario where there's no vector, i.e. vec.size() == 0 is a different condition than doesNotExist(vec). But even in that scenario, you should be preferring std::optional<std::vector> instead, which better conveys through the code the intent of the object. Granted, std::optional is only available in C++17→ Code, so maybe you're in an environment that hasn't implemented it yet. But otherwise, there's little reason to use std::unique_ptr<std::vector>.



So in general, I don't believe there are practical uses for std::unique_ptr<std::vector>. There's no practical performance difference between it and std::vector, and using it will just make your code needlessly complex.






share|improve this answer























  • Yeah, there is no need to convey optionality with unique_ptr. There are plenty of C++14 conforming implementations of optional, so one can just use one of those.

    – SergeyA
    Mar 7 at 21:53











  • Examine what happens when you have a throwing move ctor/assign in the type stored in the vector..

    – Yakk - Adam Nevraumont
    Mar 7 at 23:02











  • @Yakk-AdamNevraumont You're going to have to elaborate on what you mean, because from context, it's not clear how this is supposed to improve my answer.

    – Xirema
    Mar 8 at 0:31











  • @Xirema What happens to std::vector<TypeThatCanThrowOnAMove> when you try to move-assign it? What happens to std::unique_ptr<std::vector<TypeThatCanThrowOnAMove>>? It is a corner case, but a difference.

    – Yakk - Adam Nevraumont
    Mar 8 at 0:44












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














First of all, you shouldn't be using std::shared_ptr unless you need the specific "shared ownership" semantics associated with std::shared_ptr. If you need a smart pointer, you should default to std::unique_ptr by default, and only switch away from it in the scenario where you expressly find that you need to.



Secondly: ostensibly, the reason to prefer std::unique_ptr<TYPE> over TYPE is if you plan to move the object around a lot. This is the common design paradigm for large objects that are either unmovable, or otherwise expensive to move—i.e. they implemented a Copy Constructor and didn't implement a Move Constructor, so moves are forced to behave like a Copy.



std::vector, however, does have relatively efficient move semantics: if you move a std::vector around, regardless of how complex its contained types are, the move only constitutes a couple of pointer swaps. There's no real risk that moving a std::vector will incur a large amount of computational complexity. Even in the scenario where you're overriding a previously allocated array (invoking the Destructors of all objects in the vector), you'd still have that complexity if you were using std::unique_ptr<std::vector<TYPE>> instead, saving you nothing.



There are two advantages to std::unique_ptr<std::vector<TYPE>>. The first of which is that it gets rid of the implicit copy constructor; maybe you want to enforce to maintaining programmers that the object shouldn't be copied. But that's a pretty niche use. The other advantage is that it allows you to stipulate the scenario where there's no vector, i.e. vec.size() == 0 is a different condition than doesNotExist(vec). But even in that scenario, you should be preferring std::optional<std::vector> instead, which better conveys through the code the intent of the object. Granted, std::optional is only available in C++17→ Code, so maybe you're in an environment that hasn't implemented it yet. But otherwise, there's little reason to use std::unique_ptr<std::vector>.



So in general, I don't believe there are practical uses for std::unique_ptr<std::vector>. There's no practical performance difference between it and std::vector, and using it will just make your code needlessly complex.






share|improve this answer























  • Yeah, there is no need to convey optionality with unique_ptr. There are plenty of C++14 conforming implementations of optional, so one can just use one of those.

    – SergeyA
    Mar 7 at 21:53











  • Examine what happens when you have a throwing move ctor/assign in the type stored in the vector..

    – Yakk - Adam Nevraumont
    Mar 7 at 23:02











  • @Yakk-AdamNevraumont You're going to have to elaborate on what you mean, because from context, it's not clear how this is supposed to improve my answer.

    – Xirema
    Mar 8 at 0:31











  • @Xirema What happens to std::vector<TypeThatCanThrowOnAMove> when you try to move-assign it? What happens to std::unique_ptr<std::vector<TypeThatCanThrowOnAMove>>? It is a corner case, but a difference.

    – Yakk - Adam Nevraumont
    Mar 8 at 0:44
















2














First of all, you shouldn't be using std::shared_ptr unless you need the specific "shared ownership" semantics associated with std::shared_ptr. If you need a smart pointer, you should default to std::unique_ptr by default, and only switch away from it in the scenario where you expressly find that you need to.



Secondly: ostensibly, the reason to prefer std::unique_ptr<TYPE> over TYPE is if you plan to move the object around a lot. This is the common design paradigm for large objects that are either unmovable, or otherwise expensive to move—i.e. they implemented a Copy Constructor and didn't implement a Move Constructor, so moves are forced to behave like a Copy.



std::vector, however, does have relatively efficient move semantics: if you move a std::vector around, regardless of how complex its contained types are, the move only constitutes a couple of pointer swaps. There's no real risk that moving a std::vector will incur a large amount of computational complexity. Even in the scenario where you're overriding a previously allocated array (invoking the Destructors of all objects in the vector), you'd still have that complexity if you were using std::unique_ptr<std::vector<TYPE>> instead, saving you nothing.



There are two advantages to std::unique_ptr<std::vector<TYPE>>. The first of which is that it gets rid of the implicit copy constructor; maybe you want to enforce to maintaining programmers that the object shouldn't be copied. But that's a pretty niche use. The other advantage is that it allows you to stipulate the scenario where there's no vector, i.e. vec.size() == 0 is a different condition than doesNotExist(vec). But even in that scenario, you should be preferring std::optional<std::vector> instead, which better conveys through the code the intent of the object. Granted, std::optional is only available in C++17→ Code, so maybe you're in an environment that hasn't implemented it yet. But otherwise, there's little reason to use std::unique_ptr<std::vector>.



So in general, I don't believe there are practical uses for std::unique_ptr<std::vector>. There's no practical performance difference between it and std::vector, and using it will just make your code needlessly complex.






share|improve this answer























  • Yeah, there is no need to convey optionality with unique_ptr. There are plenty of C++14 conforming implementations of optional, so one can just use one of those.

    – SergeyA
    Mar 7 at 21:53











  • Examine what happens when you have a throwing move ctor/assign in the type stored in the vector..

    – Yakk - Adam Nevraumont
    Mar 7 at 23:02











  • @Yakk-AdamNevraumont You're going to have to elaborate on what you mean, because from context, it's not clear how this is supposed to improve my answer.

    – Xirema
    Mar 8 at 0:31











  • @Xirema What happens to std::vector<TypeThatCanThrowOnAMove> when you try to move-assign it? What happens to std::unique_ptr<std::vector<TypeThatCanThrowOnAMove>>? It is a corner case, but a difference.

    – Yakk - Adam Nevraumont
    Mar 8 at 0:44














2












2








2







First of all, you shouldn't be using std::shared_ptr unless you need the specific "shared ownership" semantics associated with std::shared_ptr. If you need a smart pointer, you should default to std::unique_ptr by default, and only switch away from it in the scenario where you expressly find that you need to.



Secondly: ostensibly, the reason to prefer std::unique_ptr<TYPE> over TYPE is if you plan to move the object around a lot. This is the common design paradigm for large objects that are either unmovable, or otherwise expensive to move—i.e. they implemented a Copy Constructor and didn't implement a Move Constructor, so moves are forced to behave like a Copy.



std::vector, however, does have relatively efficient move semantics: if you move a std::vector around, regardless of how complex its contained types are, the move only constitutes a couple of pointer swaps. There's no real risk that moving a std::vector will incur a large amount of computational complexity. Even in the scenario where you're overriding a previously allocated array (invoking the Destructors of all objects in the vector), you'd still have that complexity if you were using std::unique_ptr<std::vector<TYPE>> instead, saving you nothing.



There are two advantages to std::unique_ptr<std::vector<TYPE>>. The first of which is that it gets rid of the implicit copy constructor; maybe you want to enforce to maintaining programmers that the object shouldn't be copied. But that's a pretty niche use. The other advantage is that it allows you to stipulate the scenario where there's no vector, i.e. vec.size() == 0 is a different condition than doesNotExist(vec). But even in that scenario, you should be preferring std::optional<std::vector> instead, which better conveys through the code the intent of the object. Granted, std::optional is only available in C++17→ Code, so maybe you're in an environment that hasn't implemented it yet. But otherwise, there's little reason to use std::unique_ptr<std::vector>.



So in general, I don't believe there are practical uses for std::unique_ptr<std::vector>. There's no practical performance difference between it and std::vector, and using it will just make your code needlessly complex.






share|improve this answer













First of all, you shouldn't be using std::shared_ptr unless you need the specific "shared ownership" semantics associated with std::shared_ptr. If you need a smart pointer, you should default to std::unique_ptr by default, and only switch away from it in the scenario where you expressly find that you need to.



Secondly: ostensibly, the reason to prefer std::unique_ptr<TYPE> over TYPE is if you plan to move the object around a lot. This is the common design paradigm for large objects that are either unmovable, or otherwise expensive to move—i.e. they implemented a Copy Constructor and didn't implement a Move Constructor, so moves are forced to behave like a Copy.



std::vector, however, does have relatively efficient move semantics: if you move a std::vector around, regardless of how complex its contained types are, the move only constitutes a couple of pointer swaps. There's no real risk that moving a std::vector will incur a large amount of computational complexity. Even in the scenario where you're overriding a previously allocated array (invoking the Destructors of all objects in the vector), you'd still have that complexity if you were using std::unique_ptr<std::vector<TYPE>> instead, saving you nothing.



There are two advantages to std::unique_ptr<std::vector<TYPE>>. The first of which is that it gets rid of the implicit copy constructor; maybe you want to enforce to maintaining programmers that the object shouldn't be copied. But that's a pretty niche use. The other advantage is that it allows you to stipulate the scenario where there's no vector, i.e. vec.size() == 0 is a different condition than doesNotExist(vec). But even in that scenario, you should be preferring std::optional<std::vector> instead, which better conveys through the code the intent of the object. Granted, std::optional is only available in C++17→ Code, so maybe you're in an environment that hasn't implemented it yet. But otherwise, there's little reason to use std::unique_ptr<std::vector>.



So in general, I don't believe there are practical uses for std::unique_ptr<std::vector>. There's no practical performance difference between it and std::vector, and using it will just make your code needlessly complex.







share|improve this answer












share|improve this answer



share|improve this answer










answered Mar 7 at 21:47









XiremaXirema

14.2k11649




14.2k11649












  • Yeah, there is no need to convey optionality with unique_ptr. There are plenty of C++14 conforming implementations of optional, so one can just use one of those.

    – SergeyA
    Mar 7 at 21:53











  • Examine what happens when you have a throwing move ctor/assign in the type stored in the vector..

    – Yakk - Adam Nevraumont
    Mar 7 at 23:02











  • @Yakk-AdamNevraumont You're going to have to elaborate on what you mean, because from context, it's not clear how this is supposed to improve my answer.

    – Xirema
    Mar 8 at 0:31











  • @Xirema What happens to std::vector<TypeThatCanThrowOnAMove> when you try to move-assign it? What happens to std::unique_ptr<std::vector<TypeThatCanThrowOnAMove>>? It is a corner case, but a difference.

    – Yakk - Adam Nevraumont
    Mar 8 at 0:44


















  • Yeah, there is no need to convey optionality with unique_ptr. There are plenty of C++14 conforming implementations of optional, so one can just use one of those.

    – SergeyA
    Mar 7 at 21:53











  • Examine what happens when you have a throwing move ctor/assign in the type stored in the vector..

    – Yakk - Adam Nevraumont
    Mar 7 at 23:02











  • @Yakk-AdamNevraumont You're going to have to elaborate on what you mean, because from context, it's not clear how this is supposed to improve my answer.

    – Xirema
    Mar 8 at 0:31











  • @Xirema What happens to std::vector<TypeThatCanThrowOnAMove> when you try to move-assign it? What happens to std::unique_ptr<std::vector<TypeThatCanThrowOnAMove>>? It is a corner case, but a difference.

    – Yakk - Adam Nevraumont
    Mar 8 at 0:44

















Yeah, there is no need to convey optionality with unique_ptr. There are plenty of C++14 conforming implementations of optional, so one can just use one of those.

– SergeyA
Mar 7 at 21:53





Yeah, there is no need to convey optionality with unique_ptr. There are plenty of C++14 conforming implementations of optional, so one can just use one of those.

– SergeyA
Mar 7 at 21:53













Examine what happens when you have a throwing move ctor/assign in the type stored in the vector..

– Yakk - Adam Nevraumont
Mar 7 at 23:02





Examine what happens when you have a throwing move ctor/assign in the type stored in the vector..

– Yakk - Adam Nevraumont
Mar 7 at 23:02













@Yakk-AdamNevraumont You're going to have to elaborate on what you mean, because from context, it's not clear how this is supposed to improve my answer.

– Xirema
Mar 8 at 0:31





@Yakk-AdamNevraumont You're going to have to elaborate on what you mean, because from context, it's not clear how this is supposed to improve my answer.

– Xirema
Mar 8 at 0:31













@Xirema What happens to std::vector<TypeThatCanThrowOnAMove> when you try to move-assign it? What happens to std::unique_ptr<std::vector<TypeThatCanThrowOnAMove>>? It is a corner case, but a difference.

– Yakk - Adam Nevraumont
Mar 8 at 0:44






@Xirema What happens to std::vector<TypeThatCanThrowOnAMove> when you try to move-assign it? What happens to std::unique_ptr<std::vector<TypeThatCanThrowOnAMove>>? It is a corner case, but a difference.

– Yakk - Adam Nevraumont
Mar 8 at 0:44




















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