change for to a recursive function in c

[chrisschristou]

hello friends.
i have i simple c/c++ function that return the min of the array to the main function, in the body of the function i'm using an iterative loop so i would like to do it recursivly and i don't know from where to start! all reply are usefull thank you all Danifriends!

#include<stdio.h>
int minarr(int n, int *arr){
 int i,min;
 for(i=0; i<n; i++){
   if(min > *(arr+i)){
     min = *(arr+i);
   }
 }
 return min;
}
int main(int argc, char const *argv[]) {
 int n=10;
   int tab[10]={7,5,6,8,7,4,8,9,2,3};
 int min=minarr(n, tab);
 printf("%d", min);
 return 0;
}

[Reverend Jim]

Recursion is not an appropriate solution for this problem.

[chrisschristou]

yeah i know but would like to know how to loop thrught array using recurtion

[AssertNull]

How about this? We try not to give complete answers, but it's so short...

int minarr(int n, int *arr)
{
    if(n == 1)
        return *arr;
    int minAllButFirst = minarr(n - 1, arr+1);
    return (*arr < minAllButFirst ? *arr : minAllButFirst);
}

[Gribouillis]

I suggest this code

#include<stdio.h>

int minoffset(int n, int* arr){
    return n == 1 ? 0 : (*arr <= *(arr + (n = 1 + minoffset(n-1, arr+1))))? 0 : n;
}

int main(int argc, char const *argv[]) {
    int n=10;
    int tab[10] = {7,5,6,8,7,4,8,9,2,3};
    int min = *(tab + minoffset(n, tab));
    printf("%d\n", min);
    return 0;
}

[chrisschristou]

i know this an if condition but i don't know to use it "return n == 1 ? 0 : (arr <= *(arr + (n = 1 + minoffset(n-1, arr+1))))? 0 : n;"
"return (arr < minAllButFirst ? *arr : minAllButFirst);"

[AssertNull]

return (*arr < minAllButFirst ? *arr : minAllButFirst);

is equivalent to:

if(*arr < minAllButFirst)
    return *arr;
else
    return minAllButFirst;

*arr represents the array element at index 0.minAllButFirst represents the minimum element of indexes 1 through n-1, which is the recursive call. Return whichever is smaller. The base condition in my code is if the array size is 1, return the one and only element.