How to find the maximum number using a recursive function - c

Question

I am trying to understand how to find the maximum number using a recursive function, but I do not really understand how. Having this code:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int ar[100],n,i;
int *ptr;
printf("Enter size of the list:");
scanf("%d", &n);
printf("Printing the list:\n");
for (i = 0; i < n ; i++)
{
scanf("%d", &ar[i]);
}
ptr=&ar;
int max=maximum(ptr,n);
printf("ma %d",max);
return 0;
}
int maximum(int ar[], int n)
{
int max;
if(n+1==1)
{
return ar[n];
}
max =maximum(ar,n-1);
return ar[n]>max?ar[n]:max;
}
What is it actually doing and how?
Is it correctly using pointers to point the array of integers?
I hope you can help me understand it!

Answer 1

You set aside memory for an array of 100 integers with int ar[100], and you enter n which is set using scanf("%d", &n). If you enter a number greater than 100 at this stage, your program will seg fault because your loop for (i = 0; i < n ; i++) will try to access ar[100] which is a memory access error (the highest array index available is ar[100 - 1], notice that 100 - 1 < 100). Anyways, you fill n indices of ar in the loop. ptr = &ar just assigns the starting address of ar to ptr. By the way ptr = ar will work too, the & is not necessary. Now you can use ptr the same way you were using ar.
The easiest way to understand the recursion is to go straight to the last call of maximum. But first, understand that you passed ptr to the function which is the same as passing ar (remember, they are the same thing in main since ptr = ar.).
So in the last call to maximum, When n + 1 == 1 (same as n == 0), it returns ar[n] which is ar[0], which is first the number you entered for 'Printing the list' (it was stored in ar[0]).
Now in the second last call to maximum, n + 1 == 1 is false because n = 1 so we go to max = maximum(ar, n - 1). That's the result of the last call to maximum that I just explained, so max has the value of ar[0]. Now you have return ar[n] > max ? ar[n] : max, which is the same as return ar[1] > ar[0] ? ar[1] : ar[0]. That is the same as
if (ar[1] > ar[0]) {
return ar[1];
} else {
return ar[0];
}
And you can see that this returns whichever is bigger, ar[0] or ar[1]. Now for the third last call to maximum, max is the result of the second last call to maximum. And you can see the pattern emerge. You will return whichever is greater: max or ar[n] for all the rest of the calls to maximum, and by the time you get to the first call to maximum, you will have compared all the values in ar to find its maximum and return it.
Also, what Ajay said is right, ar[n] is accessing a value that you never initialized in your loop. You should write int max = maximum(ptr, n - 1) in main to fix that.

Answer 2

My solution with tail recursion:
int maximum(int a[], int n)
{
if (n == 1)
return a[0];
--n;
return maximum(a + (a[0] < a[n]), n);
}
Demo on compiler explorer

Answer 3

To understand how maximum works, just try and find some invariants in the maximum function
int maximum(int ar[], int n)
{
int max;
if(n==0)
{
return ar[n];
}
max =maximum(ar,n-1);
/* at this point MAX keeps the maximum of the subarray [0..N-1] and using this
we try to get by induction the maximum of [1..N]. */
return ar[n]>max?ar[n]:max;
}

Answer 4

Consider that definition :
#include <stdio.h>
#include <stdlib.h>
void maximum(int nums[], int size, int index, int * max)
{
if (index != size) {
if (nums[index] > *max)
*max = nums[index];
maximum(nums, size, ++index, max);
}
}
int main()
{
int * nums;
int size, i, max;
fprintf(stderr, "Enter size of the list: "); /* stderr to flush without \n */
if ((scanf("%d", &size) != 1) || (size <= 0)) {
puts("wrong size");
return 0;
}
if ((nums = malloc(size * sizeof(int))) == 0) {
puts("cannot allocate nums");
return 0;
}
for (i = 0; i != size; ++i) {
if (scanf("%d", &nums[i]) != 1) {
puts("invalid number");
return 0;
}
}
max = nums[0];
maximum(nums, size, 1, &max);
free(nums);
printf("max = %d\n", max);
return 0;
}
Execution:
% gcc -O2 -pedantic -Wall m.c
% ./a.out
Enter size of the list: 3
3
1
2
max = 3
maximum is recursive, but this is a terminal recursion, so the compiler can remove it to generate a loop. So maximum seems to be recursive as you request but the stack will not explode even with a large number :
% ( echo 100000000 ; yes 1 ) | ./a.out
Enter size of the list: max = 1
If I remove the option -O2 the generate code uses the recursion and the stack explode :
% gcc -pedantic -Wall m.c
% ( echo 100000000 ; yes 1 ) | ./a.out
Enter size of the list: Segmentation fault

Answer 5

int maxRecursively( arr array, int index){
int max = array.ptr[index];
if ( index == array.length-2 ){
return max= array.ptr[index]> array.ptr[index+1] ? array.ptr[index]:array.ptr[index+1];
}
return max = max > maxRecursively(array,index+1) ? max: maxRecursively(array,index+1);
}