Assign all values ​​in a dynamically allocated 2d array at one time

[ufrnkiddo]

So, I have a dynamically allocated matrix. How do I avoid having to assign the values ​​one by one? Is there a way to assign all values ​​at once?

int main (int argc, char * argv[]) 
{
	
	int **values = new int*[3];
	for (int i=0;i<3;i++)
		values[i] = new int [3];

        // How to avoid doing this...
	values[0][0] = 1;
	values[0][1] = 2;
	values[0][2] = 3;
	values[1][0] = 4;
	values[1][1] = 5;
	values[1][2] = 6;
	values[2][0] = 7;
	values[2][1] = 8;
	values[2][2] = 9;
}

[jonnin]

it depends on what the values are, and what you have.
For 1-9, yes, you can just do a for loop:

n = 1;
for(i = 0; i < 3; i++)
for(j = 0; j < 3; j++)
values[i][j] = n++;

[ufrnkiddo]

@jonnin

But this only works for values ​​that follow some pattern. What I wanted to know is if I had any way of assigning all the values ​​in a single line of code.

[jonnin]

no, there isn't. You can do patterns, or random values, but if you want to load it up with specific data, you have to assign each one, somehow, whether from a text file or user input or hard coded values... files and user input make it a loop, but you still have to type it in (file or live).

[lastchance]

#include <iostream>
using namespace std;

int main() 
{
   int **values = new int*[3];
   for ( int i = 0; i < 3; i++ ) values[i] = new int[3];

   // Moderately arbitrary set of initialisers
   int n = 0;   for ( int x : { 1, 2, 3, 40, 50, 60, 700, 800, 900 } ) { values[n/3][n%3] = x;   n++; };


   cout << "Just to check:\n";
   for ( int i = 0; i < 3; i++ )
   {
      for ( int j = 0; j < 3; j++ ) cout << i << " " << j << " | " << values[i][j] << endl;
   }
}

[JLBorges]

#include <iostream>

const std::size_t N = 3 ;
template < typename ARR_NxN > void print( const ARR_NxN& array ) ;

int main()
{
    {
        // this is the sane thing to do; we don't need dynamic storage duration
       int values[N][N] = { {1,2,3}, {4,5,6}, {7,8,9} } ;
       print(values) ;
    }

    {
        //  dynamically allocated 2d array. ideally, use std::unique_ptr<int[]>
        auto values = new int[N][N] { {1,2,3}, {4,5,6}, {7,8,9} } ;
        print(values) ;
        
        delete[] values ;
    }

    {
        // dynamically allocated array of pointers (as in the code in the earlier posts),
        // with each pointer pointing to the first element of a dynamically allocated array
        // bad: not exception safe (also as in the code in the earlier posts)
        auto values = new int* [N] { new int[N]{1,2,3}, new int[N]{4,5,6}, new int[N]{7,8,9} } ;
        print(values) ;

        for( std::size_t i = 0 ; i < N ; ++i ) delete[] values[i] ;
        delete[] values ;
    }
}

template < typename ARR_NxN > void print( const ARR_NxN& array )
{
    for( std::size_t row = 0 ; row < N ; ++row )
    {
        for( std::size_t col = 0 ; col < N ; ++col ) std::cout << array[row][col] << ' ' ;
        std::cout << '\n' ;
    }
    std::cout << "------\n" ;
}

http://coliru.stacked-crooked.com/a/208e008b27458036

[48T7M4Gy]

#include <iostream>

int main (int argc, char * argv[])
{
    int row_count = 3;
    int col_count = 3;
    
    // INITIALIZE ARRAY
    int** values = new int*[row_count];
    for(int i = 0; i < col_count; ++i)
        values[i] = new int[i];
    
    
    // DATA SOURCE
    int array[] = {1,2,3,4,5,6,7,8,9};
    
    
    // READ DATA INTO MATRIX
    for(int row = 0; row < row_count; row++){
        for(int col = 0; col < col_count; col++)
            values[row][col] = array[row*3+col];
    }
    
    
    // DISPLAY MATRIX
    for(int row = 0; row < row_count; ++row){
        for(int col = 0; col < col_count; ++col){
            std::cout << values[row][col] << ' ';
        }
        std::cout << '\n';
    }
    
    // CLEANUP GOES HERE
    
    return 0;
}

1 2 3 
4 5 6 
7 8 9 
Program ended with exit code: 0

http://stackoverflow.com/questions/936687/how-do-i-declare-a-2d-array-in-c-using-new