Consider the following C program:#include <stdio.h> void stringcopy(char *, char *); int main() { char a[30] = "@#Hello World!"; stringcopy(a, a + 2); printf("%s ", a); return 0; } void stringcopy(char *s, char *t) { while(*t) *s++ = *t++; }Which ONE of the following will be the output of the program?

Let us trace the execution of stringcopy(a, a + 2):

1. Initial contents of the array a with its 0-based indices:

   • a[0..13]: '@', '#', 'H', 'e', 'l', 'l', 'o', ' ', 'W', 'o', 'r', 'l', 'd', '!'

   • a[14]: '\0'

2. The function stringcopy(char *s, char *t) is called with s = a (points to a[0]) and t = a + 2 (points to a[2], which is 'H').

3. The while(*t) loop copies characters from *t to *s until *t points to the null terminator '\0' (a[14]):

   • a[0] <- a[2] ('H')

   • a[1] <- a[3] ('e')

   • a[2] <- a[4] ('l')

   • a[3] <- a[5] ('l')

   • a[4] <- a[6] ('o')

   • a[5] <- a[7] (' ')

   • a[6] <- a[8] ('W')

   • a[7] <- a[9] ('o')

   • a[8] <- a[10] ('r')

   • a[9] <- a[11] ('l')

   • a[10] <- a[12] ('d')

   • a[11] <- a[13] ('!')

4. At this point, t reaches a[14] which contains '\0'. The loop condition while(*t) evaluates to false and the function terminates.

5. Crucially, the function does not copy or append a null terminator '\0' to s. Therefore, indices 12 through 14 retain their original values:

   • a[12] is 'd'

   • a[13] is '!'

   • a[14] is '\0'

6. The final contents of array a up to the null terminator at index 14 are: "Hello World!d!"

Thus, `printf("%s", a);` prints Hello World!d!, matching option (D).