Assigning values to all the elements of a dynamically allocated pointer array?












4














I have a pointer array (ptrArr1) with two elements. I want the pointer array to be dynamically allocated. I can assign an address to the first element of the pointer array just fine, but I do not know how to assign an address to the second element of the pointer array. I do not wish to use any STLs or other precoded functions. I'm am doing this exercise to enhance my understanding of pointers. Thank you.



int main()

{

    int one = 1;

    int two = 2;

    int *ptrArr1 = new int[2];

    ptrArr1 = &one;

    ptrArr1[1] = &two;  //does not work

    ptrArr1 + 1 = &two; // does not work





    deleteptrArr1;

    return 0;



}





c++ arrays pointers dynamic-memory-allocation







share|improve this question




















  • 1




    When you do ptrArr1 = &one it's similar to doing one = two and expecting the value of one to still be 1.
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    Furthermore, ptrArr[x] is an int and not an int* (which e.g. &one is).
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    ptrArr[1] = &two; tries to assign an address (int*) to an int -> type mismatch. ptrArr1 + 1 = &two; is not a type mismatch but tries to assign an address to a non-lvalue. Strongly simplified, this is as wrong as 1 = 0;.
    – Scheff
    Nov 20 at 7:59












  • Also you have int *ptrArr1 = new int[2]; and right after ptrArr1 = &one; which overwrites ptrArr1. It's kind of like writing foo = 2; foo = 42;.
    – Jabberwocky
    Nov 20 at 8:27
















4














I have a pointer array (ptrArr1) with two elements. I want the pointer array to be dynamically allocated. I can assign an address to the first element of the pointer array just fine, but I do not know how to assign an address to the second element of the pointer array. I do not wish to use any STLs or other precoded functions. I'm am doing this exercise to enhance my understanding of pointers. Thank you.



int main()

{

    int one = 1;

    int two = 2;

    int *ptrArr1 = new int[2];

    ptrArr1 = &one;

    ptrArr1[1] = &two;  //does not work

    ptrArr1 + 1 = &two; // does not work





    deleteptrArr1;

    return 0;



}





c++ arrays pointers dynamic-memory-allocation







share|improve this question




















  • 1




    When you do ptrArr1 = &one it's similar to doing one = two and expecting the value of one to still be 1.
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    Furthermore, ptrArr[x] is an int and not an int* (which e.g. &one is).
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    ptrArr[1] = &two; tries to assign an address (int*) to an int -> type mismatch. ptrArr1 + 1 = &two; is not a type mismatch but tries to assign an address to a non-lvalue. Strongly simplified, this is as wrong as 1 = 0;.
    – Scheff
    Nov 20 at 7:59












  • Also you have int *ptrArr1 = new int[2]; and right after ptrArr1 = &one; which overwrites ptrArr1. It's kind of like writing foo = 2; foo = 42;.
    – Jabberwocky
    Nov 20 at 8:27














4












4








4







I have a pointer array (ptrArr1) with two elements. I want the pointer array to be dynamically allocated. I can assign an address to the first element of the pointer array just fine, but I do not know how to assign an address to the second element of the pointer array. I do not wish to use any STLs or other precoded functions. I'm am doing this exercise to enhance my understanding of pointers. Thank you.



int main()

{

    int one = 1;

    int two = 2;

    int *ptrArr1 = new int[2];

    ptrArr1 = &one;

    ptrArr1[1] = &two;  //does not work

    ptrArr1 + 1 = &two; // does not work





    deleteptrArr1;

    return 0;



}





c++ arrays pointers dynamic-memory-allocation







share|improve this question















I have a pointer array (ptrArr1) with two elements. I want the pointer array to be dynamically allocated. I can assign an address to the first element of the pointer array just fine, but I do not know how to assign an address to the second element of the pointer array. I do not wish to use any STLs or other precoded functions. I'm am doing this exercise to enhance my understanding of pointers. Thank you.



int main()

{

    int one = 1;

    int two = 2;

    int *ptrArr1 = new int[2];

    ptrArr1 = &one;

    ptrArr1[1] = &two;  //does not work

    ptrArr1 + 1 = &two; // does not work





    deleteptrArr1;

    return 0;



}





c++ arrays pointers dynamic-memory-allocation




c++ arrays pointers dynamic-memory-allocation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 at 8:14









Micha Wiedenmann

10.3k1364103




10.3k1364103










asked Nov 20 at 7:55









Carlos Robin Alvarenga

386




386








  • 1




    When you do ptrArr1 = &one it's similar to doing one = two and expecting the value of one to still be 1.
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    Furthermore, ptrArr[x] is an int and not an int* (which e.g. &one is).
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    ptrArr[1] = &two; tries to assign an address (int*) to an int -> type mismatch. ptrArr1 + 1 = &two; is not a type mismatch but tries to assign an address to a non-lvalue. Strongly simplified, this is as wrong as 1 = 0;.
    – Scheff
    Nov 20 at 7:59












  • Also you have int *ptrArr1 = new int[2]; and right after ptrArr1 = &one; which overwrites ptrArr1. It's kind of like writing foo = 2; foo = 42;.
    – Jabberwocky
    Nov 20 at 8:27














  • 1




    When you do ptrArr1 = &one it's similar to doing one = two and expecting the value of one to still be 1.
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    Furthermore, ptrArr[x] is an int and not an int* (which e.g. &one is).
    – Some programmer dude
    Nov 20 at 7:59






  • 1




    ptrArr[1] = &two; tries to assign an address (int*) to an int -> type mismatch. ptrArr1 + 1 = &two; is not a type mismatch but tries to assign an address to a non-lvalue. Strongly simplified, this is as wrong as 1 = 0;.
    – Scheff
    Nov 20 at 7:59












  • Also you have int *ptrArr1 = new int[2]; and right after ptrArr1 = &one; which overwrites ptrArr1. It's kind of like writing foo = 2; foo = 42;.
    – Jabberwocky
    Nov 20 at 8:27








1




1




When you do ptrArr1 = &one it's similar to doing one = two and expecting the value of one to still be 1.
– Some programmer dude
Nov 20 at 7:59




When you do ptrArr1 = &one it's similar to doing one = two and expecting the value of one to still be 1.
– Some programmer dude
Nov 20 at 7:59




1




1




Furthermore, ptrArr[x] is an int and not an int* (which e.g. &one is).
– Some programmer dude
Nov 20 at 7:59




Furthermore, ptrArr[x] is an int and not an int* (which e.g. &one is).
– Some programmer dude
Nov 20 at 7:59




1




1




ptrArr[1] = &two; tries to assign an address (int*) to an int -> type mismatch. ptrArr1 + 1 = &two; is not a type mismatch but tries to assign an address to a non-lvalue. Strongly simplified, this is as wrong as 1 = 0;.
– Scheff
Nov 20 at 7:59






ptrArr[1] = &two; tries to assign an address (int*) to an int -> type mismatch. ptrArr1 + 1 = &two; is not a type mismatch but tries to assign an address to a non-lvalue. Strongly simplified, this is as wrong as 1 = 0;.
– Scheff
Nov 20 at 7:59














Also you have int *ptrArr1 = new int[2]; and right after ptrArr1 = &one; which overwrites ptrArr1. It's kind of like writing foo = 2; foo = 42;.
– Jabberwocky
Nov 20 at 8:27




Also you have int *ptrArr1 = new int[2]; and right after ptrArr1 = &one; which overwrites ptrArr1. It's kind of like writing foo = 2; foo = 42;.
– Jabberwocky
Nov 20 at 8:27












2 Answers
2






active

oldest

votes


















3














You have an int array, not a pointer array. You can create a pointer array using



int **ptrArr1 = new int*[2];


and then assign the pointers to each pointer in the array:



ptrArr1[0] = &one;

ptrArr1[1] = &two;





share|improve this answer

















  • 2




    Thanks eozd! The use of '**' is completely new to me, but it lets me do what I wanted to accomplish.
    – Carlos Robin Alvarenga
    Nov 20 at 8:09










  • @CarlosRobinAlvarenga If eozd answered your question (IMHO, she/he did), please, don't forget to accept it.
    – Scheff
    Nov 20 at 8:50



















5














There is a difference between an array of integers and an array of pointers to int. In your case ptrArr1 is a pointer to an array of integers with space for two integers. Therefore you can only assign an int to ptrArr1[1] = 2 but not an address. Compare 



int xs = { 1, 2, 3 };    // an array of integers



int y0 = 42;

int *ys = { &y0, &y0 };  // an array of pointers to integers


Now you could also have pointers pointing to the first element of xs resp. ys:



int *ptr_xs = &xs[0];

int **ptr_ys = &ys[0];



// which can be simplified to:

int *ptr_xs = xs;

int **ptr_ys = ys;


For the simplification step you should look into What is array decaying?






share|improve this answer























    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53388495%2fassigning-values-to-all-the-elements-of-a-dynamically-allocated-pointer-array%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    You have an int array, not a pointer array. You can create a pointer array using



    int **ptrArr1 = new int*[2];


    and then assign the pointers to each pointer in the array:



    ptrArr1[0] = &one;
    
ptrArr1[1] = &two;





    share|improve this answer

















    • 2




      Thanks eozd! The use of '**' is completely new to me, but it lets me do what I wanted to accomplish.
      – Carlos Robin Alvarenga
      Nov 20 at 8:09










    • @CarlosRobinAlvarenga If eozd answered your question (IMHO, she/he did), please, don't forget to accept it.
      – Scheff
      Nov 20 at 8:50
















    3














    You have an int array, not a pointer array. You can create a pointer array using



    int **ptrArr1 = new int*[2];


    and then assign the pointers to each pointer in the array:



    ptrArr1[0] = &one;
    
ptrArr1[1] = &two;





    share|improve this answer

















    • 2




      Thanks eozd! The use of '**' is completely new to me, but it lets me do what I wanted to accomplish.
      – Carlos Robin Alvarenga
      Nov 20 at 8:09










    • @CarlosRobinAlvarenga If eozd answered your question (IMHO, she/he did), please, don't forget to accept it.
      – Scheff
      Nov 20 at 8:50














    3












    3








    3






    You have an int array, not a pointer array. You can create a pointer array using



    int **ptrArr1 = new int*[2];


    and then assign the pointers to each pointer in the array:



    ptrArr1[0] = &one;
    
ptrArr1[1] = &two;





    share|improve this answer












    You have an int array, not a pointer array. You can create a pointer array using



    int **ptrArr1 = new int*[2];


    and then assign the pointers to each pointer in the array:



    ptrArr1[0] = &one;
    
ptrArr1[1] = &two;






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 20 at 7:59









    eozd

    8941515




    8941515








    • 2




      Thanks eozd! The use of '**' is completely new to me, but it lets me do what I wanted to accomplish.
      – Carlos Robin Alvarenga
      Nov 20 at 8:09










    • @CarlosRobinAlvarenga If eozd answered your question (IMHO, she/he did), please, don't forget to accept it.
      – Scheff
      Nov 20 at 8:50














    • 2




      Thanks eozd! The use of '**' is completely new to me, but it lets me do what I wanted to accomplish.
      – Carlos Robin Alvarenga
      Nov 20 at 8:09










    • @CarlosRobinAlvarenga If eozd answered your question (IMHO, she/he did), please, don't forget to accept it.
      – Scheff
      Nov 20 at 8:50








    2




    2




    Thanks eozd! The use of '**' is completely new to me, but it lets me do what I wanted to accomplish.
    – Carlos Robin Alvarenga
    Nov 20 at 8:09




    Thanks eozd! The use of '**' is completely new to me, but it lets me do what I wanted to accomplish.
    – Carlos Robin Alvarenga
    Nov 20 at 8:09












    @CarlosRobinAlvarenga If eozd answered your question (IMHO, she/he did), please, don't forget to accept it.
    – Scheff
    Nov 20 at 8:50




    @CarlosRobinAlvarenga If eozd answered your question (IMHO, she/he did), please, don't forget to accept it.
    – Scheff
    Nov 20 at 8:50













    5














    There is a difference between an array of integers and an array of pointers to int. In your case ptrArr1 is a pointer to an array of integers with space for two integers. Therefore you can only assign an int to ptrArr1[1] = 2 but not an address. Compare 



    int xs = { 1, 2, 3 };    // an array of integers
    

    
int y0 = 42;
    
int *ys = { &y0, &y0 };  // an array of pointers to integers


    Now you could also have pointers pointing to the first element of xs resp. ys:



    int *ptr_xs = &xs[0];
    
int **ptr_ys = &ys[0];
    

    
// which can be simplified to:
    
int *ptr_xs = xs;
    
int **ptr_ys = ys;


    For the simplification step you should look into What is array decaying?






    share|improve this answer




























      5














      There is a difference between an array of integers and an array of pointers to int. In your case ptrArr1 is a pointer to an array of integers with space for two integers. Therefore you can only assign an int to ptrArr1[1] = 2 but not an address. Compare 



      int xs = { 1, 2, 3 };    // an array of integers
      

      
int y0 = 42;
      
int *ys = { &y0, &y0 };  // an array of pointers to integers


      Now you could also have pointers pointing to the first element of xs resp. ys:



      int *ptr_xs = &xs[0];
      
int **ptr_ys = &ys[0];
      

      
// which can be simplified to:
      
int *ptr_xs = xs;
      
int **ptr_ys = ys;


      For the simplification step you should look into What is array decaying?






      share|improve this answer


























        5












        5








        5






        There is a difference between an array of integers and an array of pointers to int. In your case ptrArr1 is a pointer to an array of integers with space for two integers. Therefore you can only assign an int to ptrArr1[1] = 2 but not an address. Compare 



        int xs = { 1, 2, 3 };    // an array of integers
        

        
int y0 = 42;
        
int *ys = { &y0, &y0 };  // an array of pointers to integers


        Now you could also have pointers pointing to the first element of xs resp. ys:



        int *ptr_xs = &xs[0];
        
int **ptr_ys = &ys[0];
        

        
// which can be simplified to:
        
int *ptr_xs = xs;
        
int **ptr_ys = ys;


        For the simplification step you should look into What is array decaying?






        share|improve this answer














        There is a difference between an array of integers and an array of pointers to int. In your case ptrArr1 is a pointer to an array of integers with space for two integers. Therefore you can only assign an int to ptrArr1[1] = 2 but not an address. Compare 



        int xs = { 1, 2, 3 };    // an array of integers
        

        
int y0 = 42;
        
int *ys = { &y0, &y0 };  // an array of pointers to integers


        Now you could also have pointers pointing to the first element of xs resp. ys:



        int *ptr_xs = &xs[0];
        
int **ptr_ys = &ys[0];
        

        
// which can be simplified to:
        
int *ptr_xs = xs;
        
int **ptr_ys = ys;


        For the simplification step you should look into What is array decaying?







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 20 at 8:21

























        answered Nov 20 at 7:59









        Micha Wiedenmann

        10.3k1364103




        10.3k1364103






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53388495%2fassigning-values-to-all-the-elements-of-a-dynamically-allocated-pointer-array%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            If I really need a card on my start hand, how many mulligans make sense? [duplicate]

            Image
            5 This question already has an answer here: How do you calculate the likelihood of drawing certain cards in your opening hand? 3 answers There is a funny modern deck where you play Treasure Hunt, but you really need to have it on the starting hand. How many mulligans should I take, if my deck contains 60 cards and contains 4 copies of Treasure Hunt? magic-the-gathering probability share | improve this question asked Nov 23 '18 at 12:12 dan dan 152 4
            Read more

            Alcedinidae

            Image
            Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
            Read more

            Can an atomic nucleus contain both particles and antiparticles? [duplicate]

            Image
            9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
            Read more