Visual Studio doesn't show full array when when dynamically creating array of pointers





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2















My code:



#include "pch.h"

#include <iostream>



using namespace std;



   int main()

   {

      char** pptr = new char*[5];



      for (int i = 0; i < 5; i++)

      pptr[i] = new char[5];

   }


What I want to happen is that pptr now points to the beginning of an array of 5 pointers that each point to the beginning of an array of 5 characters.



I put a breakpoint at the end of the main function and added pptr to watch, and it only stores one pointer. Why does this happen and how do I do it correctly?



enter image description here






c++ visual-studio pointers debugging







share|improve this question




















  • 1





    You said you wanted pptr to point to the beginning of an array of pointers... the beginning of an array is indeed one pointer - the first one in the array. The other pointers come later in the same array.

    – Galik
    Jan 4 at 14:39








  • 1





    Please next time reduce the picture to its interesting part to not have that large white margin ;-)

    – bruno
    Jan 4 at 14:44


















2















My code:



#include "pch.h"

#include <iostream>



using namespace std;



   int main()

   {

      char** pptr = new char*[5];



      for (int i = 0; i < 5; i++)

      pptr[i] = new char[5];

   }


What I want to happen is that pptr now points to the beginning of an array of 5 pointers that each point to the beginning of an array of 5 characters.



I put a breakpoint at the end of the main function and added pptr to watch, and it only stores one pointer. Why does this happen and how do I do it correctly?



enter image description here






c++ visual-studio pointers debugging







share|improve this question




















  • 1





    You said you wanted pptr to point to the beginning of an array of pointers... the beginning of an array is indeed one pointer - the first one in the array. The other pointers come later in the same array.

    – Galik
    Jan 4 at 14:39








  • 1





    Please next time reduce the picture to its interesting part to not have that large white margin ;-)

    – bruno
    Jan 4 at 14:44














2












2








2








My code:



#include "pch.h"

#include <iostream>



using namespace std;



   int main()

   {

      char** pptr = new char*[5];



      for (int i = 0; i < 5; i++)

      pptr[i] = new char[5];

   }


What I want to happen is that pptr now points to the beginning of an array of 5 pointers that each point to the beginning of an array of 5 characters.



I put a breakpoint at the end of the main function and added pptr to watch, and it only stores one pointer. Why does this happen and how do I do it correctly?



enter image description here






c++ visual-studio pointers debugging







share|improve this question
















My code:



#include "pch.h"

#include <iostream>



using namespace std;



   int main()

   {

      char** pptr = new char*[5];



      for (int i = 0; i < 5; i++)

      pptr[i] = new char[5];

   }


What I want to happen is that pptr now points to the beginning of an array of 5 pointers that each point to the beginning of an array of 5 characters.



I put a breakpoint at the end of the main function and added pptr to watch, and it only stores one pointer. Why does this happen and how do I do it correctly?



enter image description here






c++ visual-studio pointers debugging




c++ visual-studio pointers debugging






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 4 at 14:54









drescherjm

6,58923553




6,58923553










asked Jan 4 at 14:30









Viktor AxénViktor Axén

385




385








  • 1





    You said you wanted pptr to point to the beginning of an array of pointers... the beginning of an array is indeed one pointer - the first one in the array. The other pointers come later in the same array.

    – Galik
    Jan 4 at 14:39








  • 1





    Please next time reduce the picture to its interesting part to not have that large white margin ;-)

    – bruno
    Jan 4 at 14:44














  • 1





    You said you wanted pptr to point to the beginning of an array of pointers... the beginning of an array is indeed one pointer - the first one in the array. The other pointers come later in the same array.

    – Galik
    Jan 4 at 14:39








  • 1





    Please next time reduce the picture to its interesting part to not have that large white margin ;-)

    – bruno
    Jan 4 at 14:44








1




1





You said you wanted pptr to point to the beginning of an array of pointers... the beginning of an array is indeed one pointer - the first one in the array. The other pointers come later in the same array.

– Galik
Jan 4 at 14:39







You said you wanted pptr to point to the beginning of an array of pointers... the beginning of an array is indeed one pointer - the first one in the array. The other pointers come later in the same array.

– Galik
Jan 4 at 14:39






1




1





Please next time reduce the picture to its interesting part to not have that large white margin ;-)

– bruno
Jan 4 at 14:44





Please next time reduce the picture to its interesting part to not have that large white margin ;-)

– bruno
Jan 4 at 14:44












3 Answers
3






active

oldest

votes


















6














This is the default knowledge of your pointer type in Visual Studio. You indicate in the code that char** pptr is a pointer, but it cannot know how big.



To fix this, you can add a watch on pptr[0], and then you can specify that it has a "size" of 5 by changing it to pptr[0],5. Also, if the size is variable you can do "ptr[0],[size]" where size is an expression that evaluates to the number of elements to show.






share|improve this answer


























  • That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.

    – François Andrieux
    Jan 4 at 14:55













  • Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!

    – Matthieu Brucher
    Jan 4 at 14:57











  • yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)

    – bruno
    Jan 4 at 14:59













  • No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).

    – Matthieu Brucher
    Jan 4 at 14:59






  • 1





    Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.

    – François Andrieux
    Jan 4 at 15:08





















2














Your program does what you want, but the debugger cannot know the number of elements, it just know it is a pointer, so it write the contains of that pointer.



I don't know what debugger you use, but probably when you display the values you can modify char** by char*[5]` to see all






share|improve this answer































    -2














    // #include "pch.h"
    
#include <iostream>
    

    
using namespace std;
    

    
int main(){
    
    char** pptr = new char*[5];
    

    
    for (int i = 0; i < 5; i++)
    
        pptr[i] = new char[5];
    

    
    for(int i=0;i<5;i++){
    
        char ch='A';
    
        pptr[0][i]=ch;
    
    }
    

    
    for(int i=0;i<5;i++){
    
        cout<<pptr[0][i]<<" ";
    
    }
    
}


    Now the pptr[0] that is a pointer is pointing to an array of characters. Hope that it helps.






    share|improve this answer





















    • 4





      the problem doesn't concern the code but what the debugger shows by default

      – bruno
      Jan 4 at 14:50












    Your Answer






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    This is the default knowledge of your pointer type in Visual Studio. You indicate in the code that char** pptr is a pointer, but it cannot know how big.



    To fix this, you can add a watch on pptr[0], and then you can specify that it has a "size" of 5 by changing it to pptr[0],5. Also, if the size is variable you can do "ptr[0],[size]" where size is an expression that evaluates to the number of elements to show.






    share|improve this answer


























    • That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.

      – François Andrieux
      Jan 4 at 14:55













    • Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!

      – Matthieu Brucher
      Jan 4 at 14:57











    • yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)

      – bruno
      Jan 4 at 14:59













    • No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).

      – Matthieu Brucher
      Jan 4 at 14:59






    • 1





      Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.

      – François Andrieux
      Jan 4 at 15:08


















    6














    This is the default knowledge of your pointer type in Visual Studio. You indicate in the code that char** pptr is a pointer, but it cannot know how big.



    To fix this, you can add a watch on pptr[0], and then you can specify that it has a "size" of 5 by changing it to pptr[0],5. Also, if the size is variable you can do "ptr[0],[size]" where size is an expression that evaluates to the number of elements to show.






    share|improve this answer


























    • That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.

      – François Andrieux
      Jan 4 at 14:55













    • Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!

      – Matthieu Brucher
      Jan 4 at 14:57











    • yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)

      – bruno
      Jan 4 at 14:59













    • No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).

      – Matthieu Brucher
      Jan 4 at 14:59






    • 1





      Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.

      – François Andrieux
      Jan 4 at 15:08
















    6












    6








    6







    This is the default knowledge of your pointer type in Visual Studio. You indicate in the code that char** pptr is a pointer, but it cannot know how big.



    To fix this, you can add a watch on pptr[0], and then you can specify that it has a "size" of 5 by changing it to pptr[0],5. Also, if the size is variable you can do "ptr[0],[size]" where size is an expression that evaluates to the number of elements to show.






    share|improve this answer















    This is the default knowledge of your pointer type in Visual Studio. You indicate in the code that char** pptr is a pointer, but it cannot know how big.



    To fix this, you can add a watch on pptr[0], and then you can specify that it has a "size" of 5 by changing it to pptr[0],5. Also, if the size is variable you can do "ptr[0],[size]" where size is an expression that evaluates to the number of elements to show.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Jan 4 at 15:02









    SoronelHaetir

    7,1831514




    7,1831514










    answered Jan 4 at 14:37









    Matthieu BrucherMatthieu Brucher

    17.8k52445




    17.8k52445













    • That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.

      – François Andrieux
      Jan 4 at 14:55













    • Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!

      – Matthieu Brucher
      Jan 4 at 14:57











    • yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)

      – bruno
      Jan 4 at 14:59













    • No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).

      – Matthieu Brucher
      Jan 4 at 14:59






    • 1





      Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.

      – François Andrieux
      Jan 4 at 15:08





















    • That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.

      – François Andrieux
      Jan 4 at 14:55













    • Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!

      – Matthieu Brucher
      Jan 4 at 14:57











    • yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)

      – bruno
      Jan 4 at 14:59













    • No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).

      – Matthieu Brucher
      Jan 4 at 14:59






    • 1





      Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.

      – François Andrieux
      Jan 4 at 15:08



















    That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.

    – François Andrieux
    Jan 4 at 14:55







    That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.

    – François Andrieux
    Jan 4 at 14:55















    Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!

    – Matthieu Brucher
    Jan 4 at 14:57





    Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!

    – Matthieu Brucher
    Jan 4 at 14:57













    yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)

    – bruno
    Jan 4 at 14:59







    yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)

    – bruno
    Jan 4 at 14:59















    No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).

    – Matthieu Brucher
    Jan 4 at 14:59





    No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).

    – Matthieu Brucher
    Jan 4 at 14:59




    1




    1





    Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.

    – François Andrieux
    Jan 4 at 15:08







    Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.

    – François Andrieux
    Jan 4 at 15:08















    2














    Your program does what you want, but the debugger cannot know the number of elements, it just know it is a pointer, so it write the contains of that pointer.



    I don't know what debugger you use, but probably when you display the values you can modify char** by char*[5]` to see all






    share|improve this answer




























      2














      Your program does what you want, but the debugger cannot know the number of elements, it just know it is a pointer, so it write the contains of that pointer.



      I don't know what debugger you use, but probably when you display the values you can modify char** by char*[5]` to see all






      share|improve this answer


























        2












        2








        2







        Your program does what you want, but the debugger cannot know the number of elements, it just know it is a pointer, so it write the contains of that pointer.



        I don't know what debugger you use, but probably when you display the values you can modify char** by char*[5]` to see all






        share|improve this answer













        Your program does what you want, but the debugger cannot know the number of elements, it just know it is a pointer, so it write the contains of that pointer.



        I don't know what debugger you use, but probably when you display the values you can modify char** by char*[5]` to see all







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 4 at 14:34









        brunobruno

        14.4k31426




        14.4k31426























            -2














            // #include "pch.h"
            
#include <iostream>
            

            
using namespace std;
            

            
int main(){
            
    char** pptr = new char*[5];
            

            
    for (int i = 0; i < 5; i++)
            
        pptr[i] = new char[5];
            

            
    for(int i=0;i<5;i++){
            
        char ch='A';
            
        pptr[0][i]=ch;
            
    }
            

            
    for(int i=0;i<5;i++){
            
        cout<<pptr[0][i]<<" ";
            
    }
            
}


            Now the pptr[0] that is a pointer is pointing to an array of characters. Hope that it helps.






            share|improve this answer





















            • 4





              the problem doesn't concern the code but what the debugger shows by default

              – bruno
              Jan 4 at 14:50
















            -2














            // #include "pch.h"
            
#include <iostream>
            

            
using namespace std;
            

            
int main(){
            
    char** pptr = new char*[5];
            

            
    for (int i = 0; i < 5; i++)
            
        pptr[i] = new char[5];
            

            
    for(int i=0;i<5;i++){
            
        char ch='A';
            
        pptr[0][i]=ch;
            
    }
            

            
    for(int i=0;i<5;i++){
            
        cout<<pptr[0][i]<<" ";
            
    }
            
}


            Now the pptr[0] that is a pointer is pointing to an array of characters. Hope that it helps.






            share|improve this answer





















            • 4





              the problem doesn't concern the code but what the debugger shows by default

              – bruno
              Jan 4 at 14:50














            -2












            -2








            -2







            // #include "pch.h"
            
#include <iostream>
            

            
using namespace std;
            

            
int main(){
            
    char** pptr = new char*[5];
            

            
    for (int i = 0; i < 5; i++)
            
        pptr[i] = new char[5];
            

            
    for(int i=0;i<5;i++){
            
        char ch='A';
            
        pptr[0][i]=ch;
            
    }
            

            
    for(int i=0;i<5;i++){
            
        cout<<pptr[0][i]<<" ";
            
    }
            
}


            Now the pptr[0] that is a pointer is pointing to an array of characters. Hope that it helps.






            share|improve this answer















            // #include "pch.h"
            
#include <iostream>
            

            
using namespace std;
            

            
int main(){
            
    char** pptr = new char*[5];
            

            
    for (int i = 0; i < 5; i++)
            
        pptr[i] = new char[5];
            

            
    for(int i=0;i<5;i++){
            
        char ch='A';
            
        pptr[0][i]=ch;
            
    }
            

            
    for(int i=0;i<5;i++){
            
        cout<<pptr[0][i]<<" ";
            
    }
            
}


            Now the pptr[0] that is a pointer is pointing to an array of characters. Hope that it helps.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Jan 4 at 14:43









            François Andrieux

            16.7k32950




            16.7k32950










            answered Jan 4 at 14:40









            Prabhakar JhaPrabhakar Jha

            61




            61








            • 4





              the problem doesn't concern the code but what the debugger shows by default

              – bruno
              Jan 4 at 14:50














            • 4





              the problem doesn't concern the code but what the debugger shows by default

              – bruno
              Jan 4 at 14:50








            4




            4





            the problem doesn't concern the code but what the debugger shows by default

            – bruno
            Jan 4 at 14:50





            the problem doesn't concern the code but what the debugger shows by default

            – bruno
            Jan 4 at 14:50


















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            Monofisismo

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            Monofisismo (do grego: monos - "único, singular" e physis - "natureza") é o ponto de vista cristológico que defende que, depois da união do divino e do humano na encarnação histórica, Jesus Cristo, como encarnação do Filho ou Verbo ( Logos ) de Deus, teria apenas uma única "natureza", a divina, e não uma síntese de ambas. O monofisismo é contraposto pelo diofisismo (ou "diafisismo"), que defende que Jesus preservou em si as duas naturezas. Historicamente, o monofisismo se refere primordialmente à posição dos que (especialmente no Egito e, em menor grau, na Síria) rejeitaram as decisões do Concílio de Calcedônia em 451 (o quarto concílio ecumênico). Os membros mais moderados entre eles, porém, defendem a teologia "miafisita", que se tornou a oficial para as Igrejas Ortodoxas Orientais. Muitos ortodoxos orientais, porém, rejeitam essa classificação, mesmo como um termo genérico, mas ele é amplamente utilizado na literatura históri...
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            compose and upload a new article using a custom form

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            -5 1 I'm starting to learn about php trying to achieve the goal of make my first custom form. From what I understand, it's possible to collect data from a form you can compose and customize to get information like title, subtitle, uploaded images and of course the main text of a supposed post and collect them in an html file that the form itself plus a php file can store in a folder on my server. what i would like to understand better is if i can compose the form making it be able to include the collected information between pieces of code i don't wanna write anytime in the fields of the form. i also need to understand how to make it be able to save the final html file with a different suffix like an increasing number at the end of the name to be sure no older file created this way is going to be lost...
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            Mossoró

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              Nota: Para outros significados, veja Mossoró (desambiguação). Município de Mossoró "Capital do Oeste Potiguar" [ 1 ] "Capital nacional do Semiárido" [ 2 ] [ 3 ] "Capital Cultural do Rio Grande do Norte" [ 1 ] "Terra de Santa Luzia" [ 4 ] "Terra do Sol, do Sal e do Petróleo" [ 5 ] "Terra da Liberdade [ 6 ] " Centro de Mossoró, com a Praça Vigário Antônio Joaquim e, ao fundo, a câmara municipal de vereadores (esquerda) e Catedral de Santa Luzia (direita) Bandeira Brasão Hino Fundação 15 de março de 1852 (166 anos) Gentílico mossoroense Padroeiro(a) Santa Luzia Prefeito(a) Rosalba Ciarlini Rosado (PP) (2017 – 2020 ) Localização Localização de Mossoró no Rio Grande do Norte Mossoró Localização de Mossoró no Brasil 05° 11' 16" S 37° 20' 38" O 05° 11' 16" S 37° 20' 38" O Unidade federativa Rio Grande do No...
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