Group sum of n consecutive numbers of array - c++

Question

So I have an array like {1, 4, 2, 3, 5, 3, 7} and I have to make a new array containing the sums of each 3 consecutive elements like {7, 9, 10, 11, 15}.
So far I got here and I don't know what I'm doing wrong.
#include <iostream>
#include <conio.h>
using namespace std;
int a[] = {1, 1, 2, 3, 5, 3, 7};
int lung = sizeof(a)/sizeof(a[0]);
int *l = new int[10];
int calc(int *a, int m)
{
int sum = 0;
int stmax = (lung - m) - 1;
for(int st=0;st <= stmax; st++)
{
for(int i = 0; i < m; i++)
{
sum = sum + a[st+i];
}
l[st] = sum;
}
return 0;
}
void main()
{
int a[] = {1, 1, 2, 3, 5, 3, 7};
cout << calc(a, 3)<< endl;
for (int i = 0; i < lung; i++)
{
cout << l[i] << " | ";
}
_getch();
}

Answer 1

you have to set sum=0; after each iteration
for(int st=0;st <= stmax; st++)
{
sum=0;
for(int i = 0; i < m; i++)
{
sum = sum + a[st+i];
}
l[st] = sum;
}
and also you have to declare int stmax = (lung - m); //ommit -1
and in the main function print the array till stmax
for (int i = 0; i <= stmax; i++)
{
cout << l[i] << " | ";
}

Answer 2

I would do it like this:
#include <iostream>
#include <conio.h>
using namespace std;
int a[] = { 1,4,2,3,5,3,7};
int lung = sizeof(a) / sizeof(a[0]);
int *l = new int[10];
void calc(int* a, int m){
a += 1; // Increment pointer
for (int st = 0; st <m; st++){
l[st] = a[st-1]+a[st]+a[st+1];
}
}
int main(){
calc(a, lung-2);
for (auto i = 0; i < lung-2; i++){
std::cout << l[i] << " ";
}
}