please can someone tell me while the code, when run, prints 1 instead of 3.5 as solution

Question

please can someone tell me why these code, when run prints 1 instead of 3.5 to the console. it implemnts passing an array to a function, to calculate average of the array elements. i used the sizeof(array)/sizeof(int) as array length

#include "stdafx.h"

#include <iostream>

// function avg, to calculate average of an integer array.
double avg(int iArray[])
{
    // initialize variable y to calculate sum
    // initialize variable to calculate length of array    
    int y{}, z = sizeof(iArray) / sizeof(int);

    // loop over array elements
    for (int i{}; i < z; i++)
    {
        y += iArray[i];
    }

    // return average to caller
    return ((double)y) / z;
}

int main()
{
    using namespace std;

    // initialize array.
    int iArray[] = { 1, 2, 3, 4, 5, 6 };

    // call avg function, and assign return value to variable z.
    double z = avg(iArray);

    // print average to console
    cout << z << endl;

    return 0;
}

The array parameter doesn't have information about the length. `sizeof()` gives you the size of a pointer. — πάντα ῥεῖ, Oct 14 '15 at 16:04
Can an someone tell me why the behaviour is not obvious with even the most trivial debugging effort? — Martin James, Oct 14 '15 at 16:07

score 0 · Answer 1 · answered Oct 14 '15 at 16:09

double avg(int iArray[])
{
    int y{}, z = sizeof(iArray) / sizeof(int);

is equivalent to

double avg(int* iArray)
{
    int y{}, z = sizeof(int*) / sizeof(int);

You need to pass the size of the array from main.

double avg(int iArray[], int z)
{
    // initialize variable y to calculate sum
    int y{};

    // loop over array elements
    for (int i{}; i < z; i++)
    {
        y += iArray[i];
    }

    // return average to caller
    return ((double)y) / z;
}

and in main:

double z = avg(iArray, sizeof(iArray)/sizeof(iArray[0]);

please can someone tell me while the code, when run, prints 1 instead of 3.5 as solution

1 Answers1