Consider the following ANSI C program:
#include <stdio.h>
#include <stdlib.h>
struct Node{
int value;
struct Node *next; };
int main(){
struct Node *boxE, *head, *boxN; int index = 0;
boxE = head = (struct Node *) malloc(sizeof(struct Node));
head->value = index;
for (index = 1; index <= 3; index++){
boxN = (struct Node *) malloc(sizeof(struct Node));
boxE->next = boxN;
boxN->value = index;
boxE = boxN; }
for (index = 0; index <= 3; index++) {
printf("Value at index %d is %d\n", index, head->value);
head = head->next;
printf("Value at index %d is %d\n", index+1, head->value); } } Which one of the statements below is correct about the program?
Step 1: Understand the linked-list construction.
The program allocates the first node and assigns it to both \texttt{boxE} and \texttt{head}. It then creates three more nodes inside the first \texttt{for} loop (for \texttt{index = 1} to \texttt{3}) and links them using the \texttt{next} pointer.
Thus, a total of four nodes are created, not five.
Step 2: Identify the issue with the \texttt{next pointer.}
After the loop finishes, the \texttt{next} pointer of the last node is never initialized. That is, there is no statement like:
\[
\texttt{boxE->next = NULL;}
\]
As a result, the last node's \texttt{next} pointer contains an indeterminate (garbage) value.
Step 3: Analyze the second loop.
In the second \texttt{for} loop, the statement:
\[
\texttt{head = head->next;}
\]
is executed repeatedly. Eventually, when \texttt{head} points to the last node, \texttt{head->next} refers to an uninitialized pointer. Dereferencing it in:
\[
\texttt{head->value}
\]
can lead to undefined behavior or a run-time error (such as a segmentation fault).
Step 4: Eliminate incorrect options.
(A) Incorrect: Only four nodes are created.
(B) Incorrect: There is no infinite loop; both loops have fixed bounds.
(C) Incorrect: In C, reaching the end of \texttt{main} without a \texttt{return} is allowed (implicitly returns 0).
Step 5: Conclusion.
The program dereferences an uninitialized pointer, which may cause a run-time error.
Final Answer: (D)
Consider the following code:
main() {
int x = 126, y = 105;
{
if (x > y)
x = x - y;
else
y = y - x;
}
while (x != y)
printf("%d", x);
}
Consider the following code:
int a;
int arr[] = {30, 50, 10};
int *ptr = arr[10] + 1;
a = *ptr;
(*ptr)++;
ptr = ptr + 1;
printf("%d", a + arr[1] + *ptr);