Accessing arrays

Replies

• 

  Whats In Name

  xxxabhash007

  I want to know how the 2-D arrays are stored in the memory, exactly how does the compiler look at the 2-D array?😒

  Suppose you have declared a array A[4][6],
  Then in memory there will be 6 continuous blocks of 4 continuous rows allocated to this.
  
  [-][-][-][-][-]
  [-][-][-][-][-][-]
  [-][-][-][-][-][-]
  [-][-][-][-][-][-]
  
  So if you want to access S,then the address will be A[1][5].
  
  -Correct me if I am wrong.
• 

  xxxabhash007

  But I have read in Yashwant's Pointers In C:In memory there are no rows and columns. In memory whether it's a 1-D or 2-D array the elements are stored in one continuous chain.
• 

  Whats In Name

  Memory is in continuous chain,but the way to access it is different.
  It can be possible that rows are in continuous form,Eg :
  Continuous Memory:
  [-][-][-][-][-][-][-][-][-][-][-][-][-][-][-][-]
  Can be viewed as:
  Row 1:[-][-][-][-]Row 2:[-][-][-][-]Row 3:[-][-][-][-]Row 4:[-][-][-][-]So on..
  Within computer.
• 

  xxxabhash007

  What is the output?
  show(int(*q)[4],int row,int col)
  {
  int i,j,*p;
  for(i=0;i {
  p=q+i;
  for(j=0;j printf("%d",*(p+j));
  printf("\n");
  }
  printf("\n");
  }
  where argument *q[4] is receiving the array:
  a[3][4]={ {1,2,3,4},
  {5,6,7,8},
  {9,0,1,6}
  }
  row=3 & col=4
• 

  xxxabhash007

  Can anyone give me the output of the program which I have posted.
• 

  anandkumarjha

  as far as i know the output will be the address of the 3rd row and 4the column element i.e the address of the element 6....if i am wrong then plsss correct me
• 

  xxxabhash007

  Can you please give the actual output in double quotes.
• 

  Manish Goyal

  It will give you error
  
  ie you cannot convert int *[4] to int *
• 

  anandkumarjha

  goyal420

  It will give you error
  
  ie you cannot convert int *[4] to int *

  can you please explain the logic behind it
• 

  xxxabhash007

  @Goyal: There is no such kind of statement "int *[4]" or "int *". Then what are you talking about?
• 

  Manish Goyal

  show(int(*q)[4],int row,int col)
  {
   int i,j,*p;
   for(i=0;iHere p=q+i is invalid statement
  
  
  
  Since p is a pointer to array
  
  where as q is pointer to array of columns
  
   
  
  you cannot directly store address of q in p 
  
  
  
  in that case you have to modify this statement as follow
  
  p=&q[0];
  
  
  
  with this modification the  code will work and will print the array q as its output