CODEWITHSUNDEEP
c++
D1X
D1X

Reputation: 5444

Converting array of bool to array of int

I have following struct:

 struct nodo {
 array<array<bool,19>,19> tablero; 
 array<int,2> p1,p2;     
 int d1,d2;                             
 int u1,u2;                             
 } mys;

which is passed by argument to a function

void myf(nodo mys){
// that attempts to do the following conversion:
array<array<int,19>,19> lab=mys.tablero;
}

And I am receiving the following error:

error: no match for ‘operator=’ in ‘lab = mys.nodo::tablero’

So I assume the conversion cannot be done like that. What is the most efficient way to do it?

Upvotes: 0

Views: 1292

Answers (3)

erol yeniaras
erol yeniaras

Reputation: 3795

Here is how you can assign a 2D boolean array to a 2D integer array (of size 19x19 as in your case):

for(int i=0; i<19; i++) {
    for(int j=0; j<19; j++) {
       lab[i][j] = (tablero[i][j])?1:0;
    }
}

Please note the use of ternary operator in the assignment statement. If the 

tablero[i][j]

is true then 

lab[i][j] will be 1, otherwise it will be 0.

Of course you may assign different integer values as well.

Hope that helps!

Upvotes: 2

D1X
D1X

Reputation: 5444

Well, so the most straightforward method is simply:

  for (i=0; i<=18; i++){
        for (j=0; j<=18 ; j++){
                lab[i][j]= mys.tablero[i][j];
            }
 }

As Mark Ransom suggested in the first place

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

These two arrays

array<array<bool,19>,19> tablero

and

array<array<int,19>,19> lab;

have different types and there is no implicit conversion from one array to another.

You can either write loops yourself or use some standard algorithms as it is shown in this demonstrative program

#include <iostream>
#include <array>
#include <algorithm>
#include <numeric>

int main() 
{

    std::array<std::array<bool,19>,19> tablero; 
    std::array<std::array<int,19>,19> tablero1;

    std::accumulate( tablero.begin(), tablero.end(), tablero1.begin(),
                     []( auto it, const auto &a ) 
                     { 
                        return std::copy( a.begin(), a.end(), it->begin() ), ++it;
                     } );

    return 0;
}

Your compiler has to support specifier auto in lambda expressions that the code would compile.

Or the same program but with some output

#include <iostream>
#include <iomanip>
#include <array>
#include <algorithm>
#include <numeric>

int main() 
{
    const size_t N = 3;
    std::array<std::array<bool, N>, N> tablero = 
    {
        {
            { true, false, false },
            { false, true, false },
            { false, false, true }
        }
    }; 
    std::array<std::array<int, N>, N> tablero1;

    std::accumulate( tablero.begin(), tablero.end(), tablero1.begin(),
                     []( auto it, const auto &a ) 
                     { 
                        return std::copy( a.begin(), a.end(),it->begin() ), ++it;
                     } );

    for ( const auto &a : tablero )
    {
        for ( auto b : a ) std::cout << std::boolalpha << b << ' ';
        std::cout << std::endl;
    }

    std::cout << std::endl;

    for ( const auto &a : tablero1 )
    {
        for ( auto x : a ) std::cout << x << ' ';
        std::cout << std::endl;
    }

    return 0;
}


true false false 
false true false 
false false true 

1 0 0 
0 1 0 
0 0 1

Upvotes: 2

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