CODEWITHSUNDEEP
c++carrays
Luka Rusadze LukaRuso
Luka Rusadze LukaRuso

Reputation: 29

Sum computed but not stored. Why?

I've started learning functions in C today. My issue in the function "sum" where I += numbers from arrays, it does write the sum of the numbers but when the program comes out of that "sum" function the integer where I save the sum of these numbers resets to 0 and I can't seem to figure out why. This is the code, I sure do hope that you can help me out.

#include <iostream>
#include <time.h>
#include <stdlib.h>
using namespace std;
void arr(int hello[], int size) {
    srand(time(NULL));
    for (int i = 0; i < size; i++) {
        hello[i] = rand() % 100;
    }
}
void print(int hello[], int size) {
    for (int i = 0; i < size; i++) {
        cout << hello[i] << " ";
    }
}
int sum(int hello[], int size, int sumup) {
    for (int i = 0; i < size; i++) {
        sumup += hello[i];
    }
    cout << endl;
    cout << "Sum of these numbers: " << sumup << endl;
    return sumup;
}
int SumEvenOrOdd(int sumup, int size) {
    int average = sumup / size;
    cout << "Average of the all numbers in array: " << average << endl;
    return average;
}
int main() {
    bool bye = false;
    const int size = 10;
    int hello[size], sumup = 0, input;
    arr(hello, size);
    print(hello, size);
    sum(hello, size, sumup);
    SumEvenOrOdd(sumup, size);
    cin.get();
    cin.get();
    return 0;
}

Upvotes: 0

Views: 136

Answers (5)

John Bode
John Bode

Reputation: 123458

The problem is that the formal parameter sumup in the sum function is a different object from the actual parameter sumup in main, so any changes to sumup in the function are not reflected in sumup in main.

There are several ways to get around this:

  1. In C++, you can define the sumup parameter in sum to be a reference to the sumup variable in main:
    int sum( int hello[], int size, int &sumup )
  {
    // leave everything else the same.  
  }
  2. In both C and C++, you can define the sumup parameter in sum to be a pointer to the sumup variable in main:
    int sum( int hello[], int size, int *sumup ) // leading * is necessary
{
  ...
  *sumup += hello[i]; // leading * is necessary
}

int main( )
{
  ...
  sum( hello, size, &sumup ); // leading & is necessary
  ...
}
  3. Assign the result of sum back to sumup or to a different variable:
    int newsum = sum( hello, size, sumup );
    although then the question becomes, why pass sumup as an argument in the first place? Just declare a local variable within sum to hold the value, and return that:
    int sum( int hello[], int size )
{
  int result = 0;
  for ( int i = 0; i < size; ++i )
    result += hello[i];
  return result;
}

int main( )
{
  ...
  int sumup = sum( hello, size );
  ...
}

Upvotes: 2

MilConDoin
MilConDoin

Reputation: 745

Change your line sum(hello, size, sumup); to sumup = sum(hello, size, sumup);. Then your return value will be saved in the variable sumup.

If you don't want to save your sumup variable via the return value, you should learn about call by reference (and how it differs to call by value). Then you could write to the function parameter sumup and change the original value instead of the local copy inside the function.

This will open the can of worms of pointer manipulation. One of the biggest sources of FUN with C ;-)

Example: https://www.codingunit.com/c-tutorial-call-by-value-or-call-by-reference

Upvotes: 1

RoiHatam
RoiHatam

Reputation: 896

The function sum returns a value without a variable that will accept it. change this line:

   sum(hello, size, sumup);

to:

   int my_sum = sum(hello, size, sumup);

Then you will be able to use my_sum in your main.

Another solution can be passing sumup to your function by reference, just change the decleration to:

 void sum(int hello[], int size, int & sumup)

Upvotes: 0

Prajwal
Prajwal

Reputation: 4000

Try this,

sumup = sum(hello, size, sumup);

You should get right sumup value.

Upvotes: 0

Omar Martinez
Omar Martinez

Reputation: 708

Pass the function arguments as references so they can be accessed:

void sum(int hello[], int size, int& sizeup)

this will allow sizeup to retain it's value outside of the function since you're passing a reference of sizeup.

Upvotes: 0

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