Consider the following C functions.int fun1(int n) { static int i = 0; if (n > 0) { ++i; fun1(n-1); } return(i); } int fun2(int n) { static int i = 0; if (n > 0) { i = i + fun1(n); fun2(n-1); } return(i); }The return value of fun2(5) is _________.

Both fun1 and fun2 maintain their own separate static int i, initialized once to 0.

### Behavior of fun1(n): fun1(n) increments its static variable i exactly $n$ times during recursive calls when invoked with argument $n$. Since i is static and retains its value between function calls:

• Before any call, fun1's static $i = 0$.

• Calling fun1(k) increments its static $i$ by $k$ and returns the updated total value of $i$.

### Step-by-step execution of fun2(5): Let $i_1$ denote the static variable of fun1 and $i_2$ denote the static variable of fun2. Initially, $i_1 = 0$ and $i_2 = 0$.

1. fun2(5) ($n=5$):

   • fun1(5) is called: $i_1$ increases by $5 \implies i_1 = 5$, returns $5$.

   • $i_2 = i_2 + \text{fun1}(5) = 0 + 5 = 5$.

   • Calls fun2(4).

2. fun2(4) ($n=4$):

   • fun1(4) is called: $i_1$ increases by $4 \implies i_1 = 5 + 4 = 9$, returns $9$.

   • $i_2 = 5 + 9 = 14$.

   • Calls fun2(3).

3. fun2(3) ($n=3$):

   • fun1(3) is called: $i_1$ increases by $3 \implies i_1 = 9 + 3 = 12$, returns $12$.

   • $i_2 = 14 + 12 = 26$.

   • Calls fun2(2).

4. fun2(2) ($n=2$):

   • fun1(2) is called: $i_1$ increases by $2 \implies i_1 = 12 + 2 = 14$, returns $14$.

   • $i_2 = 26 + 14 = 40$.

   • Calls fun2(1).

5. fun2(1) ($n=1$):

   • fun1(1) is called: $i_1$ increases by $1 \implies i_1 = 14 + 1 = 15$, returns $15$.

   • $i_2 = 40 + 15 = 55$.

   • Calls fun2(0).

6. fun2(0) ($n=0$):

   • Condition $(n > 0)$ is false. Returns $i_2 = 55$.

All pending recursive calls unwind and return $i_2 = 55$.

Thus, the final return value is 55.