Checking which of the modules is the closest - c++

Question

Welcome. My problem is that I have given an array of numbers which I need to calculate the average (that part I did), but then I have to find the array element (module), which is closer to the average. Below paste the code (a form of main () imposed)
#include <iostream>
using namespace std;
double* aver(double* arr, size_t size, double& average){
double count;
for(int p = 0; p < size; p++)
count += arr[p];
count /= size;
double * pointer;
pointer = &count;
average = *pointer;
}
int main() {
double arr[] = {1,2,3,4,5,7};
size_t size = sizeof(arr)/sizeof(arr[0]);
double average = 0;
double* p = aver(arr,size,average);
cout << p << " " << average << endl;
}
The program should give a result
4 3.66667
I have no idea how to check which element is nearest to another, and substitute it into *p
I will be very grateful for any help.

Answer 1

Okay, this is not the answer to your problem, since you already got couple of them
How about trying something new ?
Use std::accumulate, std::sort and std::partition to achieve same goal.
#include<algorithm>
//...
struct comp
{
double avg;
comp(double x):avg(x){}
bool operator()(const double &x) const
{
return x < avg;
}
};
std::sort(arr,arr+size);
average =std::accumulate(arr, arr+size, 0.0) / size;
double *p= std::partition(arr, arr+size, comp(average));
std::cout<<"Average :"<<average <<" Closest : "<<*p<<std::endl;

Answer 2

This algorithm is based on the fact that std::map keeps its elements sorted (using operator<):
#include <map>
#include <iostream>
#include <math.h>
using namespace std;
double closest_to_avg(double* arr, size_t size, double avg) {
std::map<double,double> disturbances;
for(int p = 0; p < size; p++) {
disturbances[fabs(avg-arr[p])]=arr[p]; //if two elements are equally
} //distant from avg we take
return disturbances.begin()->second; //a new one
}

Answer 3

Since everybody is doing the kids homework...
#include <iostream>
using namespace std;
double min(double first, double second){
return first < second ? first : second;
}
double abs(double first){
return 0 < first ? first : -first;
}
double* aver(double* arr, size_t size, double& average){
double count;
for(int p = 0; p < size; p++)
count += arr[p];
average = count/size;
int closest_index = 0;
for(int p = 0; p < size; p++)
if( abs(arr[p] - average) <
abs(arr[closest_index] - average) )
closest_index = p;
return &arr[closest_index];
}
int main() {
double arr[] = {1,2,3,4,5,7};
size_t size = sizeof(arr)/sizeof(arr[0]);
double average = 0;
double* p = aver(arr,size,average);
cout << *p << " " << average << endl;
//Above ^^ gives the expected behavior,
//Without it you'll get nothing but random memory
}
I insist that you need the * before the p, it gives the value that the pointer is pointing too. Without the * then the value is the address of the memory location, which is indeterminate in this case. Ask your professor/teacher whether the specification is correct, because it isn't.
Try and understand the style and functions involved - it isn't complicated, and writing like this can go a long ways to making your graders job easier.
Also that interface is a very leaky one, in real work - consider some of the standard library algorithms and containers instead.