Sorting array based on index of minimum value? - c

Question

I'm currently trying to learn C, and the exercise I found online has me creating a function that returns the index of the smallest value in an array. This is my function:
int return_index_of_minimum(int A[10], int i, int j){
int minimum_value = A[i];
int index_to_return = 0;
for (int index = i; index < j; index++){
if (A[index] < minimum_value){
minimum_value = A[index];
index_to_return = index;
}
}
return index_to_return;
}
i and j are the lower and upper bound numbers the function should look in. For example, if i is 4 and j is 8, that means the function will return the index of the smallest value between indices 4 and 8.
Here is my main function:
#include <stdio.h>
int main(){
int numbers[10];
int user_input = 0;
for (int i = 0; i < 10; i++){
printf("Please enter a number: ");
scanf_s("%d", &user_input);
numbers[i] = user_input;
}
for (int i = 0; i < 10; i++){
int index_of_min_value = return_index_of_minimum(numbers, i, 10);
int old_num = numbers[index_of_min_value];
int new_num = numbers[i];
numbers[index_of_min_value] = new_num;
new_array[i] = old_num;
}
for (int i = 0; i < 10; i++){
printf("%d\n", new_array[i]);
}
}
The user would first enter a bunch of numbers and that would populate the array with the user's values. The idea is to use return_index_of_minimum to return the index of the smallest item in an array, and then set that equal to numbers[0] with a for loop, and then numbers[1], and then so on. old_num is the lowest number in the array, at its previous index. Here, I'm trying to swap that minimum value with whatever is at numbers[i] However, when I'm done sorting through the entire array, and am printing it out, I see that 10 (when the user enters 1-10 randomly for values) is at index 0, and then the rest of the numbers are in order. Does anybody see what is wrong here?

Answer 1

Here is a fix:
int return_index_of_minimum(int A[10], int i, int j){
int minimum_value = A[i];
int index_to_return = i;
...
}
Unfortunately this code doesn't have protection of invalid arguments, but otherwise this is an answer you've been looking for.
The reason is in call index_of_minimum(a, 9, 10): the loop performs only one iteration for index = 9, and because the minimum value is already initialized to value a[9], the index_to_return is not updated due to condition check.
This is a different approach that doesn't have same issue:
int return_index_of_minimum(int A[10], int i, int j){
/* assuming i < j */
int minimum_value = A[i];
int index_to_return = i; /* First element is a candidate */
for (int index = i + 1; index < j; index++){
/* Iterate from second element */
if (A[index] < minimum_value){
minimum_value = A[index];
index_to_return = index;
}
}
return index_to_return;
}

Answer 2

I believe there is an error in your return_index_of_minimum function.
int index_to_return = 0;
The problem lies I think here as the value of index_to_return will stay 0 if you call return_index_of_minimum(numbers, 5, 10); and that numbers[5] if the actual minimum.
However why not use a simple bubble-sort like the one implemented here
/*
* C program to sort N numbers in ascending order using Bubble sort
* and print both the given and the sorted array
*/
#include <stdio.h>
#define MAXSIZE 10
int main(void)
{
int array[MAXSIZE];
int i, j, num, temp;
printf("Enter the value of num \n");
scanf("%d", &num);
printf("Enter the elements one by one \n");
for (i = 0; i < num; i++)
{
scanf("%d", &array[i]);
}
printf("Input array is \n");
for (i = 0; i < num; i++)
{
printf("%d\n", array[i]);
}
/* Bubble sorting begins */
for (i = 0; i < num; i++)
{
for (j = 0; j < (num - i - 1); j++)
{
if (array[j] > array[j + 1])
{
temp = array[j];
array[j] = array[j + 1];
array[j + 1] = temp;
}
}
}
printf("Sorted array is...\n");
for (i = 0; i < num; i++)
{
printf("%d\n", array[i]);
}
}