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\n", 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?
Answer
(D)
Explanation
The array holds the 14 characters @#Hello World! at indices 0 to 13, followed by '\0' at index 14.
stringcopy(a, a + 2) copies one character at a time from t to s, with t always two places ahead of s. Every character is therefore read before it can be overwritten, and the 12 characters Hello World! land in indices 0 to 11.
The loop stops as soon as *t is '\0', so the terminator is never copied. Indices 12 and 13 keep their old characters d!, and the string still ends at index 14, so printf prints Hello World!d!.
Answer: D