GATE 2019
September 2, 2023Programming
September 5, 2023Programming
| Question 13 |
Consider the following C functions.
int fun1 (int n) { int fun2 (int n) {
static int i = 0; static int i = 0;
if (n > 0) { if (n > 0) {
++i; i = i + fun1 (n);
fun1 (n-1); fun2 (n-1);
} }
return (i); return (i);
} } The return value of fun2 (5) is _______.
| 55 |
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
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