Why is the output different? (putting array in a function problem)

Question

I want the output to be: 1 2 2 2

But why is the output: 1 2 3 4

What's wrong with this code?

#include <iostream>
using namespace std;

int arr[] = {0};

int pluss(int ar[],int a){
ar[0]++;
cout<<ar[0]<<endl;
if(a==0){
    pluss(ar,a+1);
    pluss(ar,a+1);
    pluss(ar,a+1);
    }
}

int main() {
pluss(arr,0);
return 0;
}

EDIT: So, the "ar" is global and not local to one child function? how to make it so the "ar" is only local to one child function? I mean: the "ar" in the first pluss(ar,1) is different from the "ar" in the second pluss(ar,2)?

I recommend that you learn how to debug your programs. If you debug your program the issue should become quite obvious. — Some programmer dude
– Some programmer dude, Commented Jul 5, 2019 at 13:24

vcp · Accepted Answer · 2019-07-05 13:28:35Z

2

Your code is equivalent of :

int main() {
pluss(arr,0);
pluss(arr,1);
pluss(arr,1);
pluss(arr,1);
return 0;
}

Since each call to pluss definitely increments the array element, before printing it, expected output is 1, 2, 3, 4.

answered Jul 5, 2019 at 13:28

vcp

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2 Comments

dpp312 Over a year ago

So, the "ar" is global and not local to one child function? how to make it so the "ar" is only local to one child function? I mean: the "ar" in the first pluss(ar,1) is different from the "ar" in the second pluss(ar,2)?

vcp Over a year ago

Method signature int pluss(int ar[],int a) is equivalent of int pluss(int *ar,int a). So you are passing pointer to first element of array, so any local change in the method is always going change the original array in main. To avoid this behavior you may pass each array element as integer value.

Armali · Accepted Answer · 2019-07-05 13:49:52Z

0

how to make it so the "ar" is only local to one child function?

If you don't like to pass each array element as integer value, you could wrap the array in a struct, since structures are passed by value rather than by reference.

#include <iostream>
using namespace std;

struct s { int a[1]; } arr = {0};

int pluss(struct s ar, int a)
{
    ar.a[0]++;
    cout <<ar.a[0] <<endl;
    if (a==0)
    {
        pluss(ar, a+1);
        pluss(ar, a+1);
        pluss(ar, a+1);
    }
}

int main()
{
    pluss(arr, 0);
    return 0;
}

answered Jul 5, 2019 at 13:49

Armali

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1 Comment

aschepler Over a year ago

Or use std::array<int, 1>, which is essentially like that struct s but with additional other features.

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