Unable to understand , how next address is getting computed in array , when performing operation &arr+1...












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  • What does getting address of array variable mean?

    7 answers




I am trying to understand array behavior. Please see the code below. Size of int is 4.



int arr={10,9,8,7,6,5};



printf("nSingle array print=> n%u  ||  %u  ||  %u  ||  %u",

       singlearr, &singlearr, &singlearr + 1);


I am getting the output:



2293248  ||  2293248  ||  2293272


I understand the expressions "singlarr" and "&singlearr" but when I am doing "&singlearr + 1", why it is giving output as 2293272 which is 24 bytes after the address 2293248 (2293248+24) ?






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 Jan 1 at 5:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















  • Pointers cannot be printed with %u. You must convert them to (void *) and print them with %p.

    – Antti Haapala
    Jan 1 at 7:43
















1
















This question already has an answer here:




  • What does getting address of array variable mean?

    7 answers




I am trying to understand array behavior. Please see the code below. Size of int is 4.



int arr={10,9,8,7,6,5};



printf("nSingle array print=> n%u  ||  %u  ||  %u  ||  %u",

       singlearr, &singlearr, &singlearr + 1);


I am getting the output:



2293248  ||  2293248  ||  2293272


I understand the expressions "singlarr" and "&singlearr" but when I am doing "&singlearr + 1", why it is giving output as 2293272 which is 24 bytes after the address 2293248 (2293248+24) ?






c







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 Jan 1 at 5:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















  • Pointers cannot be printed with %u. You must convert them to (void *) and print them with %p.

    – Antti Haapala
    Jan 1 at 7:43














1












1








1


1







This question already has an answer here:




  • What does getting address of array variable mean?

    7 answers




I am trying to understand array behavior. Please see the code below. Size of int is 4.



int arr={10,9,8,7,6,5};



printf("nSingle array print=> n%u  ||  %u  ||  %u  ||  %u",

       singlearr, &singlearr, &singlearr + 1);


I am getting the output:



2293248  ||  2293248  ||  2293272


I understand the expressions "singlarr" and "&singlearr" but when I am doing "&singlearr + 1", why it is giving output as 2293272 which is 24 bytes after the address 2293248 (2293248+24) ?






c







share|improve this question

















This question already has an answer here:




  • What does getting address of array variable mean?

    7 answers




I am trying to understand array behavior. Please see the code below. Size of int is 4.



int arr={10,9,8,7,6,5};



printf("nSingle array print=> n%u  ||  %u  ||  %u  ||  %u",

       singlearr, &singlearr, &singlearr + 1);


I am getting the output:



2293248  ||  2293248  ||  2293272


I understand the expressions "singlarr" and "&singlearr" but when I am doing "&singlearr + 1", why it is giving output as 2293272 which is 24 bytes after the address 2293248 (2293248+24) ?





This question already has an answer here:




  • What does getting address of array variable mean?

    7 answers







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edited Jan 1 at 17:35







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asked Jan 1 at 4:52









infiniteLearnerinfiniteLearner

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 Jan 1 at 5:28


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Pointers cannot be printed with %u. You must convert them to (void *) and print them with %p.

    – Antti Haapala
    Jan 1 at 7:43



















  • Pointers cannot be printed with %u. You must convert them to (void *) and print them with %p.

    – Antti Haapala
    Jan 1 at 7:43

















Pointers cannot be printed with %u. You must convert them to (void *) and print them with %p.

– Antti Haapala
Jan 1 at 7:43





Pointers cannot be printed with %u. You must convert them to (void *) and print them with %p.

– Antti Haapala
Jan 1 at 7:43












1 Answer
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&arr is a pointer to an entire array. So, if we move &arr by 1 position it will point the next block of n elements.
If the array base address is b, &arr+1 will be b + (n * 4)



here, n=6 and b = 2293248
so, &arr+1 = b+(n*4) = 2293248 + (6*4) = 2293272






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    1 Answer
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    1 Answer
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    active

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    active

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    active

    oldest

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    3














    &arr is a pointer to an entire array. So, if we move &arr by 1 position it will point the next block of n elements.
    If the array base address is b, &arr+1 will be b + (n * 4)



    here, n=6 and b = 2293248
    so, &arr+1 = b+(n*4) = 2293248 + (6*4) = 2293272






    share|improve this answer




























      3














      &arr is a pointer to an entire array. So, if we move &arr by 1 position it will point the next block of n elements.
      If the array base address is b, &arr+1 will be b + (n * 4)



      here, n=6 and b = 2293248
      so, &arr+1 = b+(n*4) = 2293248 + (6*4) = 2293272






      share|improve this answer


























        3












        3








        3







        &arr is a pointer to an entire array. So, if we move &arr by 1 position it will point the next block of n elements.
        If the array base address is b, &arr+1 will be b + (n * 4)



        here, n=6 and b = 2293248
        so, &arr+1 = b+(n*4) = 2293248 + (6*4) = 2293272






        share|improve this answer













        &arr is a pointer to an entire array. So, if we move &arr by 1 position it will point the next block of n elements.
        If the array base address is b, &arr+1 will be b + (n * 4)



        here, n=6 and b = 2293248
        so, &arr+1 = b+(n*4) = 2293248 + (6*4) = 2293272







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 1 at 5:16









        SbrTaSbrTa

        8412




        8412

















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