C Programming

Question

#include <stdio.h>
int main() {
    int A[20][30];

    int i,j,*p;
    
    p = &A[0][0];
    
    for(i=0;i<20;i++) {
    
      for(j=0;j<20;j++) {
    
        *(p+20*j+i) = i*30+j;
    
      }
    
    }
    
    printf("%d\n",A[6][10] - A[6][9]); 	
}

O/P ??

Comments

Answer is given 30. I guess its wrong then?

Answer 1

printf("%d\n",A[6][10] - A[6][9]);

• $\text{A[6][10]}$ - $7$ th rwo and $11$ th column. How many elements before $\text{A[6][10]}$ in array A ? 
  • To count that, above this element, there are $6$ rows ($30*6 = 180$ $\text{elements}$) and in $7$th row we have $10$ elements before $\text{A[6][10]}$. So, total $190$ elments before $\text{A[6][10]}$ in matrix A.
  • $\Rightarrow$ *(p+190) will be the location of $\text{A[6][10]}$.
• Similarly for  $\text{A[6][9]}$, *(p+189) will be the location.
• Reason behind evaluating locations like this is row major order storage of C style arrays

Last thing to do: represent *(p+190) and *(p+189) in terms of the given assignment format i.e *(p+20*j+i)

• *(p + 190)  = *( p + 20*9 +10) $\Rightarrow$ i = 10 and j = 9 $\Rightarrow$ A[6][10] = $10*30 + 9$ = $309$
• *(p + 189)  = *( p + 20*9 + 9) $\Rightarrow$  i = 9 and j = 9  $\Rightarrow$ A[6][9] = $9*30 + 9$ = $279$
• $\Rightarrow$ A[6][10] - A[6][9] $\rightarrow$ 30.

Comments

" A[6][10] - 7 th rwo and 11 th column  "  How is this possible ?
A[0.....20][0.....30] , right ?

I think it is 6th row and 10 th coloumn itself..

correct me if wrong..

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in array a[N] ,,,a[1] is 1st row. considering C lang $0$ based indexing

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got it :)

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:)....

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Nice explanation :)