Answer / 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.

main() { int i, n; char *x = “girl”; n = strlen(x); *x = x[n]; for(i=0; i<n; ++i) { printf(“%s\n”,x); x++; } }

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