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What is the output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D ==
arr2D[0])) );
}
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Answer Posted / susie
Answer :
1
Explanation
This is due to the close relation between the arrays and
pointers. N dimensional arrays are made up of (N-1)
dimensional arrays.
arr2D is made up of a 3 single arrays that contains 3
integers each .
The name arr2D refers to the beginning of all the 3 arrays.
*arr2D refers to the start of the first 1D array (of 3
integers) that is the same address as arr2D. So the
expression (arr2D == *arr2D) is true (1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero
doesn’t change the value/meaning. Again arr2D[0] is the
another way of telling *(arr2D + 0). So the expression
(*(arr2D + 0) == arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the
result is true(1) and the same is printed.
| Is This Answer Correct ? | 3 Yes | 0 No |
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