#include <stdlib.h>
#include <stdio.h>


/*
 * Returns the maximum value of the two integer value pointers given as parameter. One of the parameters may be NULL, but not both.
 *
 * Special cases:
 *
 * 1. If one of the values is NULL, return the other value.
 * 2. If the values are equal, return either one.
 *
 */

int max(int* param1, int* param2){

if (param1 == NULL || *param2 > *param1){
    return *param2;
}
if (param2 == NULL || *param2 <= *param1){
    return *param1;
}
else return *param1;
}




/*
 * Returns a pointer to the minimum value of the two integer value pointers given as parameter.
 * In this case both, or either of the parameters may be NULL.
 *
 * Special cases:
 *
 * 1. In case of one NULL value return the other value.
 * 2. If both of the values are NULL, return NULL.
 * 3. If the values are equal, return either one.
 *
 */


int* min(int* param1, int* param2){

if (param1 == NULL || *param2 < *param1){
        return param2;
}
if (param2 == NULL || *param2 >= *param1){
        return param1;
}

if (param1 == NULL && param2 == NULL){
    return NULL;
}
else return param1;

}





/*
 * Calculates the average of num_of_items values from standard input. The input values the function reads are of type double.
 * The program returns the calculated average as a pass-by-pointer, given to it as parameter named average.
 * You can assume that the average pointer's value has been initialized properly (e.g. to 0.0).
 *
 * Special cases:
 *
 * 1. If the value of num_of_items is zero or less, the function does nothing (no average calculations). In this case DO NOT TOUCH the average pointer in any way.
 * 2. If the num_of_items pointer is NULL, the function does nothing (no average calculations). In this case DO NOT TOUCH the average pointer in any way.
 * 3. If the average pointer given as parameter is NULL the program does nothing (no average calculations).
 * 4. In case of a failed read from standard input, the program calculates the average of the values until that point.
 *
 * Note:
 *
 * Make sure that your average function can never go to a state where it divides by zero.
 *
 */

void average(int* num_of_items, double* average){

if (*num_of_items <= 0){
    return;
}
if (num_of_items == NULL){
    return;
}
if (average == NULL){
    return;
}
int temp, sum;
for (int i=0; i < *num_of_items; i++){
    scanf("%d", &temp);
    sum+=temp;
    *average = sum / *num_of_items;
}
}