2d to 1d array conversion with help of function

Question

in this code i am trying to convert a 2d array into 1d array. I have declared the 1d array inside the function newarr(). i am unable to return the address of array arr to the main function so that I can print it.this is getting resolved if I dynamically make the array inside the function newarr().why does it happen. why cannot i initialise the array with int arr[9] inside newarr().

#include<iostream>
using namespace std;
int *newarr(int a[][3]);
void printarr(int arr[]);

int main()
{ int i,j,z=0;
 int* q;
    int a[3][3];

    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            cin>>a[i][j];
        }

  q=newarr(a);
   cout<<"new array"<<endl;
printarr(q);
}

int *newarr(int a[3][3])
{ int i,j,z=0;
 int arr[9];

    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            arr[z]=a[i][j];
            z++;
        }
    }
    return arr;

}
void printarr(int *arr)
{ int i;
    for(i=0;i<9;i++)
    {
        cout<<arr[i]<<" ";
    }

}

You cannot return address of local variable. That creates dangling pointer, leading to Undefined Behavior (UB) once used. — Jarod42, May 13 '20 at 10:04
use `std::vector` or `std::array` to avoid the quirkynes of c-arrays — 463035818_is_not_an_ai, May 13 '20 at 10:06

score 0 · Answer 1 · answered May 13 '20 at 10:19

This is happing because you declared arr inside a function which means it is local to that function. Space for arr will get deleted as soon as your function returns (or exits) and hence your returned address become invalid.

On thing you try other than dynamically allocating is declaring it as static inside function

Example :-

static int arr[9];

Space for static variable doesn't get deleted when function returns.

Kimbing Ng · Answer 2 · 2020-05-13T10:31:59.220

You cannot return address of local variable.

arr is a local variable of function newarr, It is stored in stack.

You have four simple ways to fix this.

using new

int *newarr(int a[3][3]){
    int i,j,z=0;
//    int arr[9];
    int *arr = new int[9];

    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            arr[z]=a[i][j];
            z++;
        }
    }
    return arr;
}

Don't for get to free the memory or a memory leak will occur.

Pass the pointer to the function as a parameter

int *newarr(int a[3][3], int *arr){
    int i,j,z=0;

    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            arr[z]=a[i][j];
            z++;
        }
    }
    return arr;
}

and you can call the function like:

int arr[9];
newarr(a, arr);
printArray(arr);

define a global variable

int arr[9];
int *newarr(int a[3][3]){
    int i,j,z=0;

    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            arr[z]=a[i][j];
            z++;
        }
    }
    return arr;
}

define the array as a static variable

int *newarr(int a[3][3]){
    static int arr[9];
    int i,j,z=0;
    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            arr[z]=a[i][j];
            z++;
        }
    }
    return arr;
}

For more details visit: How to access a local variable from a different function using pointers?

2d to 1d array conversion with help of function

2 Answers2

using new

Pass the pointer to the function as a parameter

define a global variable

define the array as a static variable