In C, consider this function that modifies an array through a pointer. What is printed after the loop and why? #include void change(int b, int n) { int i; for (i = 0; i < n; i++) (b + 1) = (b + i) + 5; / note the constant +1 on the left / } int main() { int i, a[] = {2, 4, 6, 8, 10}; change(a, 5); for (i = 0; i <= 4; i++) printf("%d, ", a[i]); return 0; }

Introduction / Context:This question explores pointer arithmetic and the consequences of writing to a fixed index inside a loop. The left-hand side of the assignment always uses (b + 1) rather than (b + i), which repeatedly overwrites the same array element.

Given Data / Assumptions:

• Initial array a = {2, 4, 6, 8, 10}.
• Loop runs with i = 0..4 and executes the same target index b[1] in each iteration.
• Right-hand side reads different positions via (b + i).

Concept / Approach:The statement (b + 1) = (b + i) + 5 means “store into a[1] the value a[i] + 5”. As i changes, new values overwrite a[1]. All other positions remain unchanged.

Step-by-Step Solution:i = 0 → a[1] = a[0] + 5 = 2 + 5 = 7 → a = {2, 7, 6, 8, 10}.i = 1 → a[1] = a[1] + 5 = 7 + 5 = 12 → a = {2, 12, 6, 8, 10}.i = 2 → a[1] = a[2] + 5 = 6 + 5 = 11 → a = {2, 11, 6, 8, 10}.i = 3 → a[1] = a[3] + 5 = 8 + 5 = 13 → a = {2, 13, 6, 8, 10}.i = 4 → a[1] = a[4] + 5 = 10 + 5 = 15 → a = {2, 15, 6, 8, 10}.

Verification / Alternative check:A debugger or print inside the loop will show a[1] being overwritten on each pass, ending at 15.

Why Other Options Are Wrong:"7, 9, 11, 13, 15" suggests each element incremented; the code modifies only a[1]. "2 4 6 8 10" implies no change, which is false. The negative series does not match any computed state.

Common Pitfalls:Assuming (b+1) was a typo for (b+i). Read the code literally for correct behavior.

Final Answer:2, 15, 6, 8, 10