Accessing arrays

xxxabhash007 · 2010-09-09T19:00:34+00:00

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?😒

Accessing arrays

xxxabhash007

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Updated: Oct 25, 2024

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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?😒

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Whats In Name

Member • Sep 9, 2010

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.

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xxxabhash007

Member • Sep 9, 2010

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.

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Whats In Name

Member • Sep 9, 2010

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.

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xxxabhash007

Member • Sep 9, 2010

What is the output?
show(int(*q)[4],int row,int col)
{
int i,j,*p;
for(i=0;i<row;i++)
{
p=q+i;
for(j=0;j<col;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

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xxxabhash007

Member • Sep 11, 2010

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

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anandkumarjha

Member • Sep 11, 2010

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

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xxxabhash007

Member • Sep 11, 2010

Can you please give the actual output in double quotes.

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Manish Goyal

Member • Sep 11, 2010

It will give you error

ie you cannot convert int *[4] to int *

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anandkumarjha

Member • Sep 12, 2010

goyal420
It will give you error

ie you cannot convert int *[4] to int *
can you please explain the logic behind it

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xxxabhash007

Member • Sep 13, 2010

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

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Manish Goyal

Member • Sep 14, 2010
```
show(int(*q)[4],int row,int col)
{
 int i,j,*p;
 for(i=0;i<row;i++)
 {
   p=q+i;
   for(j=0;j<col;j++)
     printf("%d",*(p+j));
   printf("\n");
 }
 printf("\n");
}
```
Here 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
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