Showing 2D array with one element at the time - c++

Question

I have a 2 dimensional array filled with 0s and 1s. I have to display that array in way that:
- 0s are always shown
- 1s are shown one at the time.
It suppose to look like a maze where 0 is a wall and 1 is a current position.
How can I do that in c++?
EDIT:
I came up with a solution but maybe there is simpler one. What if I'd create copy of my _array and copy 0s and blank spaces instead of 1s to it. Then in loop I'd assign one of _array "1" to second array then display whole array and then make swap 1 back with blank space?
EDIT2:
int _tmain(int argc, _TCHAR* argv[])
{
file();
int k=0,l=0;
for(int i=0;i<num_rows;i++)
{
for(int j=0;j<num_chars;j++)
{
if(_array[i][j] == 1)
{
k=i;
l=j;
break;
}
}
}
while(1)
{
for(int i=0;i<num_rows;i++)
{
for(int j=0;j<num_chars;j++)
{
if(_array[i][j] == 0) printf("%d",_array[i][j]);
else if(_array[i][j]==1)
{
if(k==i && l==j)
{
printf("1");
}
else printf(" ");
}
l++;
if(l>num_chars) break;
}
k++;
l=0;
printf("\n");
}
k=0;
system("cls");
}
return 0;
}
I wrote something like that but still i don't know how to clear screen in right moment. Function file() reads from file to 2D array.

Answer 1

Assuming you want something like that
000000
0 0
0000 0
0 1 0
0 0000
000000
You could print a 0 whenever it occurs and a blank space if not. To handle the current position you could use two additional variables like posX, posY. Now everytime you find a 1 in your array you check if (j == posX && i = posY) and print 1 if so...
As you just need to visualize the maze at different possible positions I'd propose a simple display function. DisplayMaze(int x, int y) is printing the maze in the required format to the screen. If _array[y][x] == 1 there is also printed a single 1...
void DisplayMaze(int x, int y)
{
for (int row = 0; row < num_rows; row++)
{
for (int col = 0; col < num_chars; col++)
{
if (_array[row][col] == 0)
std::cout << "0 ";
else if (row == y && col == x)
std::cout << "1 ";
else
std::cout << " ";
}
std::cout << std::endl;
}
std::cout << std::endl;
}
In order to display all possible positions you have to iterate over all of them and check if the current position is marked with 1 in the array (otherwise displaying would't make sense)
for (int y = 0; y < num_rows; y++)
{
for (int x = 0; x < num_chars; x++)
{
if (_array[y][x] == 1)
{
DisplayMaze(x, y);
}
}
}
The output should look like:
0 0 0 0 0 0
0 1 0
0 0 0 0 0
0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 1 0
0 0 0 0 0
0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 1 0
0 0 0 0 0
0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 1 0
0 0 0 0 0
0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0
0 0 0 0 1 0
0 0
0 0 0 0 0
0 0 0 0 0 0
...
However, i'd recommend a more C++ like approach as a maze could be implemented as a class. This class could bring it's own display-method and would encapsulate the internal data. It could basically look like:
class Maze
{
public:
// generate empty maze with given size
Maze(int width, int height);
// destructor
~Maze();
// print maze if the given position is marked with 1
void printPosition(int x, int y) const;
// takes a cstring as input to initialize the maze from
Maze& operator<<(const char* input);
// returns true if the given position is marked with 1
bool isValidPosition(int x, int y) const;
private:
// this is the actual representation of the maze
std::vector<std::vector<int> > grid_;
};
it would be used as followes:
Maze myMaze(num_chars, num_rows);
myMaze << "000000"
"011110"
"000010"
"011110"
"010000"
"000000";
for (int y = 0; y < num_rows; y++)
{
for (int x = 0; x < num_chars; x++)
{
if (myMaze.isValidPosition(x,y))
{
myMaze.printPosition(x,y);
}
}
}

Answer 2

hire you go*[solved]*
#include <cstdlib>
#include <iostream>
using namespace std;
int main()
{
int x,y;
cin>>x>>y;
char map[x][y];
memset(map, 'a', sizeof(map));
int y_pos = 0;
for (int x_pos = 0; x_pos < x * y; x_pos++){
if (x_pos == x){
x_pos = 0;
y_pos = y_pos + 1;
cout<<endl;
}
if (y_pos == y){
system("pause");
return 0;
}
cout<<map[x_pos][y_pos];
}