What is the output of the following C code?
#include <stdio.h>
int fun(int x, int y) {
if (x == 0) return y;
return fun(x - 1, x + y);
}
int main(void) {
printf("%d", fun(4, 3));
return 0;
}fun(4,3) -> fun(3, 4+3=7) -> fun(2, 3+7=10) -> fun(1, 2+10=12) -> fun(0, 1+12=13) -> returns 13.
What is printed by calling print_star(3)?
void print_star(int n) {
if (n <= 0) return;
for(int i=0; i<n; i++) printf("*");
printf("
");
print_star(n - 1);
}Prints n stars, newline, then recurses with n-1: 3 stars, 2 stars, 1 star.
What is the primary risk of using deep recursion without tail call optimization?
Exceeding available call stack memory leads to stack overflow.
What is the output of the C function for call tr(5)?
int tr(int n) {
if (n <= 0) return 0;
return n % 2 + tr(n / 2);
}tr(5) = 1 + tr(2) = 1 + 0 + tr(1) = 1 + 0 + 1 + tr(0) = 2 (counts 1-bits in 5 = 101_2).
Consider the code:
int solve(int n) {
if (n <= 1) return 1;
return solve(n - 1) + solve(n - 2);
}
What is the base case condition?The termination condition preventing further recursion is n <= 1.