Btukfyl

I'm trying to read the following maze.txt file:

35

35

0

10

++++++++++S++++++++++++++++++++++++

++++++++++OOOOOOOOOOOOOOOOOOOOOOOOO

++++++++++O++++++++++++++++++O+++++

OOOOOOOOOOOOOOOOOOOOOOOOOO+++O++OOE

O+++++++++O++++++++++++++O+++O++O++

OOOOOO++++O++++++++++++++O+++O++O++

O++++O++++OOOOOOOOOOO++++O+OOO++O++

O++++O++++O+++++++++OOOO+O+++O++O++

OOO++OOOO+OOOOOO+++++++++++OOO++OOO

O+O+++++O++++++OOOOOOOOOO++O++++++O

O+OOOO++O++++++O++++++++O+++OOO+++O

O++++O++OOOOOOOO++++++++O+++O+O+++O

OOO++O++++++++++++++++++OOOOO+O+++O

++O++OOOOOOOOOOOOOOOO+++++++++OO++O

OOO+++++++++++++++++OOOOOO++++++++O

O++++++++++++++++++++++++O++OOOOOOO

+++++++++++++++++++++++++O++O++++++

OOOOOOOOOOOOOOOOOOOOOOOOOO++OOOOO++

O++++++++++++++++++++++++O++++++O++

OOOOOOO+++++++++++++++OOOOOOO+++O++

++++++++++++++++++++++O+++++OO++O++

OOOOOOOOOOOOOOOOOOOOOOO++++++O++O++

O++++++++++++++++++++++++++++O+OOOO

OOOO++++++++++++++++++++OOOOOO+O+++

+++OOOOOOOOO+++++++++++++++++++O+++

+++++O+++++OOOOOOOOOO++++++++OOO+++

+O+++OOOOO++++++O++++++++++++O+++++

+OOOOO+++O++++++OOOOOO+++++++O+++++

+++++++++++++++++++++OOOOOOOOO+++++

OOOOOOOOOOOOOOOOOOOOOO+++++++++++++

+++++++++++++++++++++++++++++++++++

+++++++++++++++++++++++++++++++++++

+++++++++++++++++++++++++++++++++++

+++++++++++++++++++++++++++++++++++

+++++++++++++++++++++++++++++++++++

The code works fine with the maze inside the code but I moved it out to a text file, which seems to be read but it is not working. It's giving me the error:

No matching function for call to 'mazeTravel'.

I'm not sure where to go from here. Any help would be appreciated!

#include <iostream>

#include <fstream>



using namespace std;



void printMaze(const char maze[12], int xCoordinate, int yCoordinate);

int mazeTravel(char maze[12], int xCoordinate, int yCoordinate, int direction);



int main()

{

    char maze[35][35];

    ifstream file;

    file.open("maze.txt");

    if (!file) {

        cout << "Error reading filen";

        return -1;

    }

    else {

        for (int row = 0; row < 35; row++) {

            for (int column = 0; column < 35; column++) {

                file >> maze[row][column];



                int success = 0;

                success = mazeTravel(maze, 2, 0, 1);

                if (success == 1)

                    cout << "The maze has been solved.n";

                else

                    cout << "Sorry, the maze cannot be solvedn";

            }

        }

    }

    return 0;

}

You could use a std::vector of std::strings to represent your maze:

std::vector<std::string> maze;

To access its cells use

maze[row][column];  // with row being y and column x

To get the number of rows use

maze.size()

and

maze[0].size()

for the number of columns.

You could read such a maze like that (without error checking to not clutter the code):

std::vector<std::string> readMaze(std::istream &is)

{

    std::size_t columns;        

    std::size_t rows;

    is >> columns >> rows;



    int foo;             // sorry, don't know what the 3rd and 4th 

    is >> foo >> foo;    // number is. a starting position, perhaps?



    // ignore the rest of the current line:

    is.ignore(std::numeric_limits<std::streamsize>::max(), 'n');



    std::string line;

    std::vector<std::string> maze;



    while (std::getline(is, line))

        maze.push_back(line);



    return maze;

}

An implementation (with error checking) could look like that:

#include <cstdlib>  // EXIT_FAILURE

#include <limits>   // std::numeric_limits<>

#include <vector>

#include <string>

#include <fstream>

#include <iostream>



// to not have to type std::vector<std::string> all over the place

using maze_type = std::vector<std::string>;



void printMazeCell(maze_type const &maze, std::size_t x, std::size_t y)

{

    std::cout.put(maze[y][x]);

}



void printMaze(maze_type const &maze)

{

    for (auto const &row : maze)

        std::cout << row << 'n';

}



int mazeTravel(maze_type const &maze, std::size_t x, std::size_t y, int dir)

{

    // access cells of the maze with maze[y][x]

    // maze.size() for the number of columns and

    // maze[0].size() for the number of rows

    return 42;

}



maze_type readMaze(std::istream &is)

{

    std::size_t columns;

    if (!(is >> columns)) {

        std::cerr << "Couldn't read the number of columns :(nn";

        return maze_type{};  // return an empty maze on error

    }



    std::size_t rows;

    if (!(is >> rows)) {

        std::cerr << "Couldn't read the number of rows  :(nn";

        return maze_type{};

    }



    int foo;

    is >> foo >> foo;  // sorry, don't know what the 3rd and 4th number is



    // ignore the rest of the current line:

    is.ignore(std::numeric_limits<std::streamsize>::max(), 'n');



    std::cout << "Trying to read a maze with " << columns << " columns ...n";



    std::string line;

    maze_type maze;

    while (std::getline(is, line)) {

        if (line.length() != columns) {

            std::cerr << "Found a row that contains only "

                << line.length() << " columns :(nn";

            return maze_type{};

        }

        maze.push_back(line);

    }



    if (maze.size() != rows) {

        std::cerr << "The maze only consists of "

                  << maze.size() << " rows :(nn";

        return maze_type{};

    }



    return maze;

}



int main()

{

    char const *filename = "maze.txt";

    std::ifstream is{ filename };

    if (!is.is_open()) {

        std::cerr << "Couldn't open "" << filename << "" for reading :(nn";

        return EXIT_FAILURE;

    }



    maze_type maze = readMaze(is);



    if (!maze.size()) {  // readMaze returned an empty maze :(

        std::cerr << "Bye.n";

        return EXIT_FAILURE;

    }



    printMaze(maze);

}

The problem that you don't have the implementation of

int mazeTravel(char maze[12], int xCoordinate, int yCoordinate, int direction);

You should create the implementation like this:

int mazeTravel(char maze[12], int xCoordinate, int yCoordinate, int direction)

{

    // The implementation

    return 0;

}

Another thing: You have to read the first some numbers at the beginning of the file.

35

35

0

10

After that you can read the matrix from the file