Consider the C function given below.int f(int j){ static int i = 50; i...
Understanding the Function Behavior
The function `f(int j)` contains a static variable `i` initialized to 50. It checks if `i` equals `j` and behaves differently based on this condition.
Function Execution Flow
- When `f` is called with `j = 50`:
- `i` is 50, so the condition `if (i == j)` is true.
- It prints "something".
- It then calls itself recursively with `k = f(i)`, which is `k = f(50)`.
Infinite Recursion Issue
- Each call to `f(50)` will meet the same condition:
- It will keep printing "something" and call `f(50)` again.
- This creates an infinite loop of function calls, as there is no exit condition for `j = 50`.
Impact on Runtime Stack
- Each recursive call consumes stack space.
- Eventually, the program will run out of stack memory, leading to a stack overflow or crash.
Other Options Explained
- Option A: The function does return 0 for all values of `j`, but this is not the main concern; the infinite loop is the critical issue.
- Option B: It only prints "something" when `j = 50`, not for all values.
- Option C: While it returns 0 when `j = 50`, this does not address the infinite recursion problem.
In conclusion, the correct answer is option D: the function will exhaust the runtime stack or run into an infinite loop when `j = 50`.
Consider the C function given below.int f(int j){ static int i = 50; i...
When j is 50, the function would call itself again and again as neither i nor j is changed inside the recursion.