Changing Array elements by call by value

Question

I am passing array elements to a function. This function adds 5 to each element of the array. I am also passing an integer and adding 5 to it... Even though it is a 'call by value' function the value of the integer dosn't change in main() (which is expected) but the array elements do change...

I wan't to know how and why?

#include <iostream>

using namespace std;

void change(int x[],int y);

int main()
{
    int sharan[]={1,2,3,4};
    int a=10;
    change(sharan,a);
    for(int j=0;j<4;j++)
    {
        cout<<sharan[j]<<endl;
    }
    cout<<endl<<"a is : "<<a;
    return(0);
}

void change(int x[],int y)
{
    for(int i=0;i<4;i++)
    {
        x[i]+=5;
    }
    y+=5;
}

When you call `change(sharan,a)` a copy of the value of a is made and used inside the function. (The function assigns 5 to that copy which is fine but has nothing to do with the outside `a` which stays unaffected.) Declare the parameter `y` as a reference like this: `void change(int x[],int &y){ ... }`. Then y will become an alias for (not a copy of) of a and changes to y will be equivalent to changes of a. — Peter - Reinstate Monica, Oct 02 '14 at 07:33

score 5 · Accepted Answer · answered Oct 02 '14 at 07:30

Array decays to pointer,

void change(int x[],int y) is equivalent to void change (int *x, int y )

with x[i] += 5;

you're changing the content of address at x+i

y=5; inside change basically updates a local copy of y whose address wasn't passed, hence no modification in actual value of y after change exists

score 1 · Answer 2 · answered Oct 02 '14 at 07:30

1

since array is always a reference type so any changes outside will affect the array in calling function too.

As you can see in your code :

change(sharan,a); // here `sharan` points the base address an you are passing it.

answered Oct 02 '14 at 07:30

Rustam

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@PaulR it is `Address` – Rustam Oct 02 '14 at 07:34

score 1 · Answer 3 · answered Oct 02 '14 at 07:40

C++ does not support passing raw arrays by value. When you try to pass them as such they are converted into a pointer to the array. For this reason if you modify the elements of the array then the change will be reflected in the calling function. In this case main().

Changing Array elements by call by value

3 Answers3