I need the drawing of the call stack and the heap Consider t
I need the drawing of the call stack and the heap
Consider the following C functions: void main(...) {int result = fib(3); printf(\"Result = %, d\ \", result);} int fib(int n) {if (nSolution
#include<stdio.h>
int main()
{
int result = fibo(7);
printf(\"\ \ Result = %d \ \", result);
}
int fibo(int n)
{
static int c, d, r;
if(n < 2)
{
printf(\"\ n = %d\\t\", n);
return n;
}
printf(\"\ fibo(n - 1) = %d\\t\", c);
printf(\"\ fibo(n - 2) = %d\\t\", d);
printf(\"\ fibo(n - 1) + fibo(n - 2) = %d\\t\", r);
return r = (fibo(n - 1) + fibo(n - 2));
return r;
}
Output:
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 0
n = 1
n = 0
n = 1
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 2
n = 1
n = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 3
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 3
n = 1
n = 0
n = 1
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 5
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 5
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 5
n = 1
n = 0
n = 1
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 2
n = 1
n = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 8
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 8
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 8
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 8
n = 1
n = 0
n = 1
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 2
n = 1
n = 0
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 3
fibo(n - 1) = 0
fibo(n - 2) = 0
fibo(n - 1) + fibo(n - 2) = 3
n = 1
n = 0
n = 1
Result = 13
if the parameter is 50.

