c++ float* array as reference to std::vector












4















I want to create a std::vector<float> vpd which will be a reference to float*.



float * old = new float[6];
for (int i = 0; i < 6; i++)
{
old[i] = i;
}
vector<float> vpd(6);
auto refasd = &*vpd.begin();
*refasd = *old;
vpd[0] = 23;
cout << old[0] << endl;


How should I modify the code, if I want get 23 from cout?










share|improve this question

























  • Are you maybe looking for this: stackoverflow.com/questions/2434196/…

    – uceumern
    Nov 21 '18 at 10:52











  • Is it about telling the vector to use your "preallocated" buffer instead of the one a vector manages internally, or is it just that you want to refer to to the vector's internal buffer in terms of a float*-type?

    – Stephan Lechner
    Nov 21 '18 at 10:52













  • maybe you meant vpd[0] = 23; *old = *refasd; ?

    – M.M
    Nov 21 '18 at 10:52
















4















I want to create a std::vector<float> vpd which will be a reference to float*.



float * old = new float[6];
for (int i = 0; i < 6; i++)
{
old[i] = i;
}
vector<float> vpd(6);
auto refasd = &*vpd.begin();
*refasd = *old;
vpd[0] = 23;
cout << old[0] << endl;


How should I modify the code, if I want get 23 from cout?










share|improve this question

























  • Are you maybe looking for this: stackoverflow.com/questions/2434196/…

    – uceumern
    Nov 21 '18 at 10:52











  • Is it about telling the vector to use your "preallocated" buffer instead of the one a vector manages internally, or is it just that you want to refer to to the vector's internal buffer in terms of a float*-type?

    – Stephan Lechner
    Nov 21 '18 at 10:52













  • maybe you meant vpd[0] = 23; *old = *refasd; ?

    – M.M
    Nov 21 '18 at 10:52














4












4








4


1






I want to create a std::vector<float> vpd which will be a reference to float*.



float * old = new float[6];
for (int i = 0; i < 6; i++)
{
old[i] = i;
}
vector<float> vpd(6);
auto refasd = &*vpd.begin();
*refasd = *old;
vpd[0] = 23;
cout << old[0] << endl;


How should I modify the code, if I want get 23 from cout?










share|improve this question
















I want to create a std::vector<float> vpd which will be a reference to float*.



float * old = new float[6];
for (int i = 0; i < 6; i++)
{
old[i] = i;
}
vector<float> vpd(6);
auto refasd = &*vpd.begin();
*refasd = *old;
vpd[0] = 23;
cout << old[0] << endl;


How should I modify the code, if I want get 23 from cout?







c++ arrays reference stdvector






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 '18 at 10:50









DeiDei

6,20153355




6,20153355










asked Nov 21 '18 at 10:39









Artur DomasikArtur Domasik

212




212













  • Are you maybe looking for this: stackoverflow.com/questions/2434196/…

    – uceumern
    Nov 21 '18 at 10:52











  • Is it about telling the vector to use your "preallocated" buffer instead of the one a vector manages internally, or is it just that you want to refer to to the vector's internal buffer in terms of a float*-type?

    – Stephan Lechner
    Nov 21 '18 at 10:52













  • maybe you meant vpd[0] = 23; *old = *refasd; ?

    – M.M
    Nov 21 '18 at 10:52



















  • Are you maybe looking for this: stackoverflow.com/questions/2434196/…

    – uceumern
    Nov 21 '18 at 10:52











  • Is it about telling the vector to use your "preallocated" buffer instead of the one a vector manages internally, or is it just that you want to refer to to the vector's internal buffer in terms of a float*-type?

    – Stephan Lechner
    Nov 21 '18 at 10:52













  • maybe you meant vpd[0] = 23; *old = *refasd; ?

    – M.M
    Nov 21 '18 at 10:52

















Are you maybe looking for this: stackoverflow.com/questions/2434196/…

– uceumern
Nov 21 '18 at 10:52





Are you maybe looking for this: stackoverflow.com/questions/2434196/…

– uceumern
Nov 21 '18 at 10:52













Is it about telling the vector to use your "preallocated" buffer instead of the one a vector manages internally, or is it just that you want to refer to to the vector's internal buffer in terms of a float*-type?

– Stephan Lechner
Nov 21 '18 at 10:52







Is it about telling the vector to use your "preallocated" buffer instead of the one a vector manages internally, or is it just that you want to refer to to the vector's internal buffer in terms of a float*-type?

– Stephan Lechner
Nov 21 '18 at 10:52















maybe you meant vpd[0] = 23; *old = *refasd; ?

– M.M
Nov 21 '18 at 10:52





maybe you meant vpd[0] = 23; *old = *refasd; ?

– M.M
Nov 21 '18 at 10:52












5 Answers
5






active

oldest

votes


















5














You can't. std::vector is not designed to take ownership of a raw pointer.

Maybe you can make do with std::unique_ptr<float>, but the better solution is to directly use std::vector.






share|improve this answer



















  • 1





    And just as I was about to hit submit..

    – StoryTeller
    Nov 21 '18 at 10:46



















5














As alternative, you might use std::span (C++20)



float* old = new float[6];
std::iota(old, old + 6, 0);
std::span<float> vpd(old, 6);
vpd[0] = 23;
std::cout << old[0] << std::endl;
delete old;





share|improve this answer





















  • 2





    Plus one from the future. Does span own the memory, i.e. is the delete intentionally missing?

    – Bathsheba
    Nov 21 '18 at 11:11











  • Span does not own the memory if I'm not mistaken, so yeah, a delete is missing.

    – rubenvb
    Nov 21 '18 at 11:18



















4














You could also create a vector of std::reference_wrapper objects that refer to the original float array - may that be std::vector<float> or a newed float*. An example:



vector<std::reference_wrapper<float>> vpd(old, old + 6); // ¹

vpd[0].get() = 23.f;

cout << old[0] << endl; // prints 23


¹) Thanks to @StoryTeller for pointing out that vpd can be directly initialized.






share|improve this answer





















  • 1





    I'm not sure you need std::transform. IIRC std::vector<std::reference_wrapper<float>> vpd(old, old + 6); should work out of the box.

    – StoryTeller
    Nov 21 '18 at 10:55











  • @StoryTeller True, that's a nice improvement!

    – lubgr
    Nov 21 '18 at 10:58



















2














As std::vector has its own memory structure you cannot map vector<float> or even vector<float*> to an array of float. However you can map each vector item to an array one.



float* old = new float[6];
for (int i = 0; i < 6; i++)
{
old[i] = i;
}
vector<float*> vpd(6);
int i = 0;
for (auto it = vpd.begin(); it != vpd.end(); it++)
{
*it = &old[i++];
}
*vpd[0] = 23;
*vpd[2] = 45;
cout << old[0] << endl << old[2] << endl;


Output



23
45





share|improve this answer


























  • You ruined your cache locality by doing this, though

    – Lightness Races in Orbit
    Nov 21 '18 at 11:17











  • @LightnessRacesinOrbit cold you please explain in more details?

    – serge
    Nov 21 '18 at 11:58











  • gameprogrammingpatterns.com/data-locality.html

    – Lightness Races in Orbit
    Nov 21 '18 at 12:47











  • @LightnessRacesinOrbit the link is cool in design context however I still don't see the relation with reusing of some memory piece accessed by pointers.

    – serge
    Nov 21 '18 at 13:25






  • 1





    Now you're talking about the cost of creating those elements, not the cost of accessing them. But the answer is still yes.

    – Lightness Races in Orbit
    Nov 21 '18 at 13:52



















0














Depending on version of C++ you'll have some options (if I understand the problem):
if you have an older c++ version then c++11 I would in this case declear std::vector as:



// then you really need to make sure u delete the memory as well
std::vector<float*> vpd(6);


If however you got c++11 or higher I would use either std::share_ptr or std::unique_ptr depending on if you would like to share the memory space or not. Either this will ensure the memory is deleted by it self without having to do a "delete float*;" which is nice.
You can read about std::shared_ptr at: https://en.cppreference.com/w/cpp/memory/shared_ptr and std::unique_ptr at: https://en.cppreference.com/w/cpp/memory/unique_ptr



// For unique_ptr
std::vector<std::unique_ptr<float>> vpd(6);
// for std::shared_ptr
std::vector<std::shared_ptr<float>> vpd(6);


I would say that if you can then use unique_ptr rather then shared_ptr due to shared_ptr has som extra komplexitity within to be make sure the memory not used before deleteing memory space.






share|improve this answer























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    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    You can't. std::vector is not designed to take ownership of a raw pointer.

    Maybe you can make do with std::unique_ptr<float>, but the better solution is to directly use std::vector.






    share|improve this answer



















    • 1





      And just as I was about to hit submit..

      – StoryTeller
      Nov 21 '18 at 10:46
















    5














    You can't. std::vector is not designed to take ownership of a raw pointer.

    Maybe you can make do with std::unique_ptr<float>, but the better solution is to directly use std::vector.






    share|improve this answer



















    • 1





      And just as I was about to hit submit..

      – StoryTeller
      Nov 21 '18 at 10:46














    5












    5








    5







    You can't. std::vector is not designed to take ownership of a raw pointer.

    Maybe you can make do with std::unique_ptr<float>, but the better solution is to directly use std::vector.






    share|improve this answer













    You can't. std::vector is not designed to take ownership of a raw pointer.

    Maybe you can make do with std::unique_ptr<float>, but the better solution is to directly use std::vector.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 21 '18 at 10:45









    QuentinQuentin

    46.4k589146




    46.4k589146








    • 1





      And just as I was about to hit submit..

      – StoryTeller
      Nov 21 '18 at 10:46














    • 1





      And just as I was about to hit submit..

      – StoryTeller
      Nov 21 '18 at 10:46








    1




    1





    And just as I was about to hit submit..

    – StoryTeller
    Nov 21 '18 at 10:46





    And just as I was about to hit submit..

    – StoryTeller
    Nov 21 '18 at 10:46













    5














    As alternative, you might use std::span (C++20)



    float* old = new float[6];
    std::iota(old, old + 6, 0);
    std::span<float> vpd(old, 6);
    vpd[0] = 23;
    std::cout << old[0] << std::endl;
    delete old;





    share|improve this answer





















    • 2





      Plus one from the future. Does span own the memory, i.e. is the delete intentionally missing?

      – Bathsheba
      Nov 21 '18 at 11:11











    • Span does not own the memory if I'm not mistaken, so yeah, a delete is missing.

      – rubenvb
      Nov 21 '18 at 11:18
















    5














    As alternative, you might use std::span (C++20)



    float* old = new float[6];
    std::iota(old, old + 6, 0);
    std::span<float> vpd(old, 6);
    vpd[0] = 23;
    std::cout << old[0] << std::endl;
    delete old;





    share|improve this answer





















    • 2





      Plus one from the future. Does span own the memory, i.e. is the delete intentionally missing?

      – Bathsheba
      Nov 21 '18 at 11:11











    • Span does not own the memory if I'm not mistaken, so yeah, a delete is missing.

      – rubenvb
      Nov 21 '18 at 11:18














    5












    5








    5







    As alternative, you might use std::span (C++20)



    float* old = new float[6];
    std::iota(old, old + 6, 0);
    std::span<float> vpd(old, 6);
    vpd[0] = 23;
    std::cout << old[0] << std::endl;
    delete old;





    share|improve this answer















    As alternative, you might use std::span (C++20)



    float* old = new float[6];
    std::iota(old, old + 6, 0);
    std::span<float> vpd(old, 6);
    vpd[0] = 23;
    std::cout << old[0] << std::endl;
    delete old;






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 21 '18 at 11:18









    Bathsheba

    180k27254383




    180k27254383










    answered Nov 21 '18 at 11:08









    Jarod42Jarod42

    118k12103189




    118k12103189








    • 2





      Plus one from the future. Does span own the memory, i.e. is the delete intentionally missing?

      – Bathsheba
      Nov 21 '18 at 11:11











    • Span does not own the memory if I'm not mistaken, so yeah, a delete is missing.

      – rubenvb
      Nov 21 '18 at 11:18














    • 2





      Plus one from the future. Does span own the memory, i.e. is the delete intentionally missing?

      – Bathsheba
      Nov 21 '18 at 11:11











    • Span does not own the memory if I'm not mistaken, so yeah, a delete is missing.

      – rubenvb
      Nov 21 '18 at 11:18








    2




    2





    Plus one from the future. Does span own the memory, i.e. is the delete intentionally missing?

    – Bathsheba
    Nov 21 '18 at 11:11





    Plus one from the future. Does span own the memory, i.e. is the delete intentionally missing?

    – Bathsheba
    Nov 21 '18 at 11:11













    Span does not own the memory if I'm not mistaken, so yeah, a delete is missing.

    – rubenvb
    Nov 21 '18 at 11:18





    Span does not own the memory if I'm not mistaken, so yeah, a delete is missing.

    – rubenvb
    Nov 21 '18 at 11:18











    4














    You could also create a vector of std::reference_wrapper objects that refer to the original float array - may that be std::vector<float> or a newed float*. An example:



    vector<std::reference_wrapper<float>> vpd(old, old + 6); // ¹

    vpd[0].get() = 23.f;

    cout << old[0] << endl; // prints 23


    ¹) Thanks to @StoryTeller for pointing out that vpd can be directly initialized.






    share|improve this answer





















    • 1





      I'm not sure you need std::transform. IIRC std::vector<std::reference_wrapper<float>> vpd(old, old + 6); should work out of the box.

      – StoryTeller
      Nov 21 '18 at 10:55











    • @StoryTeller True, that's a nice improvement!

      – lubgr
      Nov 21 '18 at 10:58
















    4














    You could also create a vector of std::reference_wrapper objects that refer to the original float array - may that be std::vector<float> or a newed float*. An example:



    vector<std::reference_wrapper<float>> vpd(old, old + 6); // ¹

    vpd[0].get() = 23.f;

    cout << old[0] << endl; // prints 23


    ¹) Thanks to @StoryTeller for pointing out that vpd can be directly initialized.






    share|improve this answer





















    • 1





      I'm not sure you need std::transform. IIRC std::vector<std::reference_wrapper<float>> vpd(old, old + 6); should work out of the box.

      – StoryTeller
      Nov 21 '18 at 10:55











    • @StoryTeller True, that's a nice improvement!

      – lubgr
      Nov 21 '18 at 10:58














    4












    4








    4







    You could also create a vector of std::reference_wrapper objects that refer to the original float array - may that be std::vector<float> or a newed float*. An example:



    vector<std::reference_wrapper<float>> vpd(old, old + 6); // ¹

    vpd[0].get() = 23.f;

    cout << old[0] << endl; // prints 23


    ¹) Thanks to @StoryTeller for pointing out that vpd can be directly initialized.






    share|improve this answer















    You could also create a vector of std::reference_wrapper objects that refer to the original float array - may that be std::vector<float> or a newed float*. An example:



    vector<std::reference_wrapper<float>> vpd(old, old + 6); // ¹

    vpd[0].get() = 23.f;

    cout << old[0] << endl; // prints 23


    ¹) Thanks to @StoryTeller for pointing out that vpd can be directly initialized.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 21 '18 at 11:00

























    answered Nov 21 '18 at 10:49









    lubgrlubgr

    13.2k31850




    13.2k31850








    • 1





      I'm not sure you need std::transform. IIRC std::vector<std::reference_wrapper<float>> vpd(old, old + 6); should work out of the box.

      – StoryTeller
      Nov 21 '18 at 10:55











    • @StoryTeller True, that's a nice improvement!

      – lubgr
      Nov 21 '18 at 10:58














    • 1





      I'm not sure you need std::transform. IIRC std::vector<std::reference_wrapper<float>> vpd(old, old + 6); should work out of the box.

      – StoryTeller
      Nov 21 '18 at 10:55











    • @StoryTeller True, that's a nice improvement!

      – lubgr
      Nov 21 '18 at 10:58








    1




    1





    I'm not sure you need std::transform. IIRC std::vector<std::reference_wrapper<float>> vpd(old, old + 6); should work out of the box.

    – StoryTeller
    Nov 21 '18 at 10:55





    I'm not sure you need std::transform. IIRC std::vector<std::reference_wrapper<float>> vpd(old, old + 6); should work out of the box.

    – StoryTeller
    Nov 21 '18 at 10:55













    @StoryTeller True, that's a nice improvement!

    – lubgr
    Nov 21 '18 at 10:58





    @StoryTeller True, that's a nice improvement!

    – lubgr
    Nov 21 '18 at 10:58











    2














    As std::vector has its own memory structure you cannot map vector<float> or even vector<float*> to an array of float. However you can map each vector item to an array one.



    float* old = new float[6];
    for (int i = 0; i < 6; i++)
    {
    old[i] = i;
    }
    vector<float*> vpd(6);
    int i = 0;
    for (auto it = vpd.begin(); it != vpd.end(); it++)
    {
    *it = &old[i++];
    }
    *vpd[0] = 23;
    *vpd[2] = 45;
    cout << old[0] << endl << old[2] << endl;


    Output



    23
    45





    share|improve this answer


























    • You ruined your cache locality by doing this, though

      – Lightness Races in Orbit
      Nov 21 '18 at 11:17











    • @LightnessRacesinOrbit cold you please explain in more details?

      – serge
      Nov 21 '18 at 11:58











    • gameprogrammingpatterns.com/data-locality.html

      – Lightness Races in Orbit
      Nov 21 '18 at 12:47











    • @LightnessRacesinOrbit the link is cool in design context however I still don't see the relation with reusing of some memory piece accessed by pointers.

      – serge
      Nov 21 '18 at 13:25






    • 1





      Now you're talking about the cost of creating those elements, not the cost of accessing them. But the answer is still yes.

      – Lightness Races in Orbit
      Nov 21 '18 at 13:52
















    2














    As std::vector has its own memory structure you cannot map vector<float> or even vector<float*> to an array of float. However you can map each vector item to an array one.



    float* old = new float[6];
    for (int i = 0; i < 6; i++)
    {
    old[i] = i;
    }
    vector<float*> vpd(6);
    int i = 0;
    for (auto it = vpd.begin(); it != vpd.end(); it++)
    {
    *it = &old[i++];
    }
    *vpd[0] = 23;
    *vpd[2] = 45;
    cout << old[0] << endl << old[2] << endl;


    Output



    23
    45





    share|improve this answer


























    • You ruined your cache locality by doing this, though

      – Lightness Races in Orbit
      Nov 21 '18 at 11:17











    • @LightnessRacesinOrbit cold you please explain in more details?

      – serge
      Nov 21 '18 at 11:58











    • gameprogrammingpatterns.com/data-locality.html

      – Lightness Races in Orbit
      Nov 21 '18 at 12:47











    • @LightnessRacesinOrbit the link is cool in design context however I still don't see the relation with reusing of some memory piece accessed by pointers.

      – serge
      Nov 21 '18 at 13:25






    • 1





      Now you're talking about the cost of creating those elements, not the cost of accessing them. But the answer is still yes.

      – Lightness Races in Orbit
      Nov 21 '18 at 13:52














    2












    2








    2







    As std::vector has its own memory structure you cannot map vector<float> or even vector<float*> to an array of float. However you can map each vector item to an array one.



    float* old = new float[6];
    for (int i = 0; i < 6; i++)
    {
    old[i] = i;
    }
    vector<float*> vpd(6);
    int i = 0;
    for (auto it = vpd.begin(); it != vpd.end(); it++)
    {
    *it = &old[i++];
    }
    *vpd[0] = 23;
    *vpd[2] = 45;
    cout << old[0] << endl << old[2] << endl;


    Output



    23
    45





    share|improve this answer















    As std::vector has its own memory structure you cannot map vector<float> or even vector<float*> to an array of float. However you can map each vector item to an array one.



    float* old = new float[6];
    for (int i = 0; i < 6; i++)
    {
    old[i] = i;
    }
    vector<float*> vpd(6);
    int i = 0;
    for (auto it = vpd.begin(); it != vpd.end(); it++)
    {
    *it = &old[i++];
    }
    *vpd[0] = 23;
    *vpd[2] = 45;
    cout << old[0] << endl << old[2] << endl;


    Output



    23
    45






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 21 '18 at 11:56

























    answered Nov 21 '18 at 11:00









    sergeserge

    70148




    70148













    • You ruined your cache locality by doing this, though

      – Lightness Races in Orbit
      Nov 21 '18 at 11:17











    • @LightnessRacesinOrbit cold you please explain in more details?

      – serge
      Nov 21 '18 at 11:58











    • gameprogrammingpatterns.com/data-locality.html

      – Lightness Races in Orbit
      Nov 21 '18 at 12:47











    • @LightnessRacesinOrbit the link is cool in design context however I still don't see the relation with reusing of some memory piece accessed by pointers.

      – serge
      Nov 21 '18 at 13:25






    • 1





      Now you're talking about the cost of creating those elements, not the cost of accessing them. But the answer is still yes.

      – Lightness Races in Orbit
      Nov 21 '18 at 13:52



















    • You ruined your cache locality by doing this, though

      – Lightness Races in Orbit
      Nov 21 '18 at 11:17











    • @LightnessRacesinOrbit cold you please explain in more details?

      – serge
      Nov 21 '18 at 11:58











    • gameprogrammingpatterns.com/data-locality.html

      – Lightness Races in Orbit
      Nov 21 '18 at 12:47











    • @LightnessRacesinOrbit the link is cool in design context however I still don't see the relation with reusing of some memory piece accessed by pointers.

      – serge
      Nov 21 '18 at 13:25






    • 1





      Now you're talking about the cost of creating those elements, not the cost of accessing them. But the answer is still yes.

      – Lightness Races in Orbit
      Nov 21 '18 at 13:52

















    You ruined your cache locality by doing this, though

    – Lightness Races in Orbit
    Nov 21 '18 at 11:17





    You ruined your cache locality by doing this, though

    – Lightness Races in Orbit
    Nov 21 '18 at 11:17













    @LightnessRacesinOrbit cold you please explain in more details?

    – serge
    Nov 21 '18 at 11:58





    @LightnessRacesinOrbit cold you please explain in more details?

    – serge
    Nov 21 '18 at 11:58













    gameprogrammingpatterns.com/data-locality.html

    – Lightness Races in Orbit
    Nov 21 '18 at 12:47





    gameprogrammingpatterns.com/data-locality.html

    – Lightness Races in Orbit
    Nov 21 '18 at 12:47













    @LightnessRacesinOrbit the link is cool in design context however I still don't see the relation with reusing of some memory piece accessed by pointers.

    – serge
    Nov 21 '18 at 13:25





    @LightnessRacesinOrbit the link is cool in design context however I still don't see the relation with reusing of some memory piece accessed by pointers.

    – serge
    Nov 21 '18 at 13:25




    1




    1





    Now you're talking about the cost of creating those elements, not the cost of accessing them. But the answer is still yes.

    – Lightness Races in Orbit
    Nov 21 '18 at 13:52





    Now you're talking about the cost of creating those elements, not the cost of accessing them. But the answer is still yes.

    – Lightness Races in Orbit
    Nov 21 '18 at 13:52











    0














    Depending on version of C++ you'll have some options (if I understand the problem):
    if you have an older c++ version then c++11 I would in this case declear std::vector as:



    // then you really need to make sure u delete the memory as well
    std::vector<float*> vpd(6);


    If however you got c++11 or higher I would use either std::share_ptr or std::unique_ptr depending on if you would like to share the memory space or not. Either this will ensure the memory is deleted by it self without having to do a "delete float*;" which is nice.
    You can read about std::shared_ptr at: https://en.cppreference.com/w/cpp/memory/shared_ptr and std::unique_ptr at: https://en.cppreference.com/w/cpp/memory/unique_ptr



    // For unique_ptr
    std::vector<std::unique_ptr<float>> vpd(6);
    // for std::shared_ptr
    std::vector<std::shared_ptr<float>> vpd(6);


    I would say that if you can then use unique_ptr rather then shared_ptr due to shared_ptr has som extra komplexitity within to be make sure the memory not used before deleteing memory space.






    share|improve this answer




























      0














      Depending on version of C++ you'll have some options (if I understand the problem):
      if you have an older c++ version then c++11 I would in this case declear std::vector as:



      // then you really need to make sure u delete the memory as well
      std::vector<float*> vpd(6);


      If however you got c++11 or higher I would use either std::share_ptr or std::unique_ptr depending on if you would like to share the memory space or not. Either this will ensure the memory is deleted by it self without having to do a "delete float*;" which is nice.
      You can read about std::shared_ptr at: https://en.cppreference.com/w/cpp/memory/shared_ptr and std::unique_ptr at: https://en.cppreference.com/w/cpp/memory/unique_ptr



      // For unique_ptr
      std::vector<std::unique_ptr<float>> vpd(6);
      // for std::shared_ptr
      std::vector<std::shared_ptr<float>> vpd(6);


      I would say that if you can then use unique_ptr rather then shared_ptr due to shared_ptr has som extra komplexitity within to be make sure the memory not used before deleteing memory space.






      share|improve this answer


























        0












        0








        0







        Depending on version of C++ you'll have some options (if I understand the problem):
        if you have an older c++ version then c++11 I would in this case declear std::vector as:



        // then you really need to make sure u delete the memory as well
        std::vector<float*> vpd(6);


        If however you got c++11 or higher I would use either std::share_ptr or std::unique_ptr depending on if you would like to share the memory space or not. Either this will ensure the memory is deleted by it self without having to do a "delete float*;" which is nice.
        You can read about std::shared_ptr at: https://en.cppreference.com/w/cpp/memory/shared_ptr and std::unique_ptr at: https://en.cppreference.com/w/cpp/memory/unique_ptr



        // For unique_ptr
        std::vector<std::unique_ptr<float>> vpd(6);
        // for std::shared_ptr
        std::vector<std::shared_ptr<float>> vpd(6);


        I would say that if you can then use unique_ptr rather then shared_ptr due to shared_ptr has som extra komplexitity within to be make sure the memory not used before deleteing memory space.






        share|improve this answer













        Depending on version of C++ you'll have some options (if I understand the problem):
        if you have an older c++ version then c++11 I would in this case declear std::vector as:



        // then you really need to make sure u delete the memory as well
        std::vector<float*> vpd(6);


        If however you got c++11 or higher I would use either std::share_ptr or std::unique_ptr depending on if you would like to share the memory space or not. Either this will ensure the memory is deleted by it self without having to do a "delete float*;" which is nice.
        You can read about std::shared_ptr at: https://en.cppreference.com/w/cpp/memory/shared_ptr and std::unique_ptr at: https://en.cppreference.com/w/cpp/memory/unique_ptr



        // For unique_ptr
        std::vector<std::unique_ptr<float>> vpd(6);
        // for std::shared_ptr
        std::vector<std::shared_ptr<float>> vpd(6);


        I would say that if you can then use unique_ptr rather then shared_ptr due to shared_ptr has som extra komplexitity within to be make sure the memory not used before deleteing memory space.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 21 '18 at 10:58









        MikaelMikael

        111




        111






























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