C++ 2D array to 1D array

Question

I am attempting to convert a 2D array to 1D. I'm extremely new to C/C++ but I think it's very important to learn how to convert a 2D array to 1D. So here I am stumbling upon this problem.

My code so far is http://ideone.com/zvjKwP

#include<iostream>

using namespace std;

int main()
{

int n=0,m=0; // 2D array nRow, nCol
int a[n][m];
int i,j; // цикъл въвеждане 2D
int q,p,t; // for 2D=>1D
int b[100];
int r; // for cout
cout<<"Enter the array's number of rows and columns: ";
cin>>n>>m;

// entering values for the 2D array
    for (i = 0;i<=n;i++)
    {
        for (j = 0;j<=m;j++)
        {
            cout<<"a["<<i<<"]["<<j<<"]="<<endl;
            cin>>a[i][j];
            cin.ignore();
        }
    }

  // Most likely the failzone IMO
  for (q = 0;q<=i;q++)
    {
        for (t = 0;t<=i*j+j;t++)
        {
            b[t] = a[i][j];
        }
    }
    // attempting to print the 1D values
     cout<<"The values in the array are"<<"\n";
    for(r=0;r<=t;r++)
    {
        cout<<"b["<<r<<"] = "<<b[r]<<endl;
    }

    cin.get();
    return 0;
    }

I wrote a comment at where I think I fail.

I must also limit the numbers that get into the 1D to numbers who's value^2 is greater than 50. But for sure I must solve the problem with the conversion 2D=>1D Can you help me out?

Answer 1

http://www.cplusplus.com/doc/tutorial/arrays/

In that link look at the section on pseudo-multidimensional arrays.

I've seen many examples that that get the subscripting algorithm wrong. If in doubt, trace it out. The order of sub-scripting a 2D array should go sequentially from 0-(HEIGHT*WIDTH-1)

#define WIDTH 5
#define HEIGHT 3

int jimmy [HEIGHT * WIDTH];
int n,m;

int main ()
{
  for (n=0; n<HEIGHT; n++)
    for (m=0; m<WIDTH; m++)
    {
      jimmy[n*WIDTH+m]=(n+1)*(m+1);
    }
}

Answer 2

You are right with your supposition:

The cycle should be like:

for (q = 0; q < n; q++)
{
    for (t = 0; t < m; t++)
    {
        b[q * m + t] = a[q][t];
    }
}

It is always easier to consider such conversions from the view point of the higher dimension array. Furthermore with your code you did not actually modify i or j in the b assigning cycle, so you should not expect different values to be assigned to the different array members of b .

Answer 3

This code

int n=0,m=0; // 2D array nRow, nCol
int a[n][m];

is invalid. First of all the dimensions shall be constant expressions and there is no sense to set them to 0.

And the more simple way to do your task is to use pointer. For example

int *p = b;

for ( const auto &row : a )
{
    for ( int x : row ) *p++ = x;
}

Answer 4

You can also do it this way;

int singleArraySize = columns * rows;
for (int i = 0; i < singleArraySize; ++i)
    *(oneDArr + i) = *(twoDArr + i);

the second example is taking advantage of fact that 2D array occupies consecutive spaces.

So if you have array[3][3] = {{1,2,3}, {4,5,6}, {7,8,9}} in computer memory the are stored like this:

memory address | array[i][j] | value
---------------+-------------+---------
0x1            | array[0][0] | 1
0x2            | array[0][1] | 2
0x3            | array[0][2] | 3
0x4            | array[1][0] | 4
0x5            | array[1][1] | 5
0x6            | array[1][2] | 6
0x7            | array[2][0] | 7
0x8            | array[2][1] | 8
0x9            | array[2][2] | 9

where memory address is address where value is stored.

Answer 5

First of all, the size of the 1D array should be n*m .

The cycle can be as follows-

int lim = n*m;

for(q = 0; q<lim; ++q) {

    b[q] = a[q/m][q%m];
}

Answer 6

This answer uses the C utility memcpy . Since memory is contiguous, you can copy it as a block of bytes:

#include <string.h> // For memcpy
#include <iostream>

int main() {
    // Fill in the 2D array with some values...
    int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}};

    // Allocate the 1D array
    int b[3*3];

    // Copy into 1D array
    memcpy(b,a,9*sizeof(int));

    // Print the result
    for (int i(0); i<9; ++i) {
        std::cout << b[i] << " ";
    }
}

Output:

1 2 3 4 5 6 7 8 9