c++ - Comparing the sums of the elements on both sides of an element 'N' in an array(attempt 2) -

i took consideration hints received, applied modular thinking , tried again. program runs. depending on set of values hard-wire elements of array, receive output, index sum of elements on left equal sum of elements on right. understand have been objective of exercise.

i chose not use vector in exercise because need practice remembering array has constant pointer position 1 , such, when array passed function, 1 must remember pass along size of array, or,alternately inside function array being passed, 1 can loop through array , count number of elements therein, thereafter using count array size.

please criticize new , functional code , point out else have done wrong. thank you.

#include "stdafx.h" #include <iostream> using namespace std;  /*************************************** * right side of array * calculates sum of elements right of n ***************************************/ int rightsideofarray(int arrayone[], int size, int i) {     int n = 0;      //loop through array , right hand sum     (int j = 1 + i; j < size; j++)     {          n += arrayone[j];      }     return n;   }   /*************************************** * left side of array * calculates sum of elements left of n ***************************************/ int leftsideofarray(int arrayone[], int size, int i) {     int n2 = 0;      //find left hand sum     (int j = - 1; j >= 0; j--)     {         n2 += arrayone[j];      }     return n2;  }    int main() {     //define , declare array     int const size = 7;     int arrayone[size] =     { 1,2,3,4,3,2,1 };     int n = 0;     int n2 = 0;     int count = 0;      //do comparison     (int = 0; < size; i++)     {      //compare right hand , left hand side , return right values     if (rightsideofarray(arrayone, size, i) ==      leftsideofarray(arrayone,       size, i))       counter++;     cout << << endl;  }     if (counter == 0)      cout << -1 << endl;  system("pause");  return 0;  } 

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old code: first attempt read previous solution same query can't figure out went wrong. challenge understand loop through integer array, @ each, element 'i', must add elements left of 'i' 'left side sum'. must add elements right of 'i' 'right hand sum'. there after, should compare sums right hand , left hand sides of array. if both sums equal, should have function return index @ equalization of right hand , left hand side occurred. else, should return -1.

can tell me why getting '-1' answer?

int equalsidesofanarray(int arrayone[], int n, int n2) {      //loop through array , right hand sum     (int = 0; < sizeof(arrayone); i++)     {          (int j = 1 + i; j < sizeof(arrayone); j++)         {             n += arrayone[j];             n2 += arrayone[j - 1];         }          if (n == n2)             return arrayone[i];         else             return -1;      }  }  int main() {     // define , declare array     int const size = 7;     int arrayone[size] = { 1, 2, 3, 4, 3, 2, 1 };     int n = 0;     int n2 = 0;       int answer = equalsidesofanarray(arrayone, n, n2);     cout << answer << endl;       system("pause");      return 0;  } 

first of all, arrayone parameter of function pointer first element of array, , sizeof(arrayone) size of pointer, not size size of array.
, within main(), sizeof(arrayone) return size * sizeof(int).

as coding in c++, use std::vector/std::array , banish c arrays. save such trouble , much more.

and think initializing n , n2 (which don't need pass parameter), , returning -1.