Pointers to pointers: add one statement inside fun so that j in main receives the address of local a (demonstration of double-pointer assignment). #include int main() { int j; void fun(int*); fun(&j); return 0; } void fun(int *k) { int a = 10; / Add a statement here */ }

Introduction / Context:This exercise demonstrates how to write through a pointer-to-pointer parameter in order to modify a pointer variable in the caller. The function fun receives the address of j (type int) so it can assign a target address into j.

Given Data / Assumptions:

• Main declares int j; then calls fun(&j); therefore inside fun, k points to j.
• Local variable in fun: int a = 10; located on the stack of fun.
• Goal: store the address of a into j via k.

Concept / Approach:To change a caller’s pointer through a function parameter, pass a pointer to that pointer. Then assign through the double-pointer: k = &a; This writes the address value into j. Note: this demonstrates mechanics but leaves j dangling after fun returns because a is a local variable.

Step-by-Step Solution:k is int; k is the caller’s int (j).&a is the address of the local int a.Assign *k = &a; which sets j to point at a.After fun returns, a ceases to exist → j becomes a dangling pointer (unsafe to dereference).

Verification / Alternative check:To avoid a dangling pointer, allocate a dynamically: int *p = malloc(sizeof *p); *p = 10; *k = p; and later free in the caller.

Why Other Options Are Wrong:**k = a assigns an int into a pointer variable. k = &a changes only the local parameter, not j, and also types mismatch. &k = *a is illegal. *k = a assigns value 10 to j (type mismatch).

Common Pitfalls:Confusing the levels of indirection; ignoring lifetime of local variables when storing their addresses into caller pointers.

Final Answer:*k = &a;