Programming

int fun1(int n) {
printf(“–fun1 call–\n”);
static int i = 0;
if(n>0){
++i;
printf(“fun1(%d-1)\n”,n);
fun1(n-1);
}
printf(“fun1(%d)= %d\n”,n, i);
return(i);
}

int fun2(int n) {
printf(“\n******* fun2 call ********\n”);
static int i = 0;
if(n>0){
printf(“%d + fun1(%d)\n”, i,n);
i=i+fun1(n);
fun2(n-1);
}
printf(“fun2(%d)= %d\n”,n, i);
return(i);
}

void main()
{
printf(“final = %d\n”, fun2(5));
}

Check step by step hand run of the code to understand the recursion:
******* fun2 call ********
0 + fun1(5)
–fun1 call–
fun1(5-1)
–fun1 call–
fun1(4-1)
–fun1 call–
fun1(3-1)
–fun1 call–
fun1(2-1)
–fun1 call–
fun1(1-1)
–fun1 call–
fun1(0)= 5
fun1(1)= 5
fun1(2)= 5
fun1(3)= 5
fun1(4)= 5
fun1(5)= 5

******* fun2 call ********
5 + fun1(4)
–fun1 call–
fun1(4-1)
–fun1 call–
fun1(3-1)
–fun1 call–
fun1(2-1)
–fun1 call–
fun1(1-1)
–fun1 call–
fun1(0)= 9
fun1(1)= 9
fun1(2)= 9
fun1(3)= 9
fun1(4)= 9

******* fun2 call ********
14 + fun1(3)
–fun1 call–
fun1(3-1)
–fun1 call–
fun1(2-1)
–fun1 call–
fun1(1-1)
–fun1 call–
fun1(0)= 12
fun1(1)= 12
fun1(2)= 12
fun1(3)= 12

******* fun2 call ********
26 + fun1(2)
–fun1 call–
fun1(2-1)
–fun1 call–
fun1(1-1)
–fun1 call–
fun1(0)= 14
fun1(1)= 14
fun1(2)= 14

******* fun2 call ********
40 + fun1(1)
–fun1 call–
fun1(1-1)
–fun1 call–
fun1(0)= 15
fun1(1)= 15
******* fun2 call ********
fun2(0)= 55
fun2(1)= 55
fun2(2)= 55
fun2(3)= 55
fun2(4)= 55
fun2(5)= 55
final = 55