CODEWITHSUNDEEP
carraysmaze
Justin Thurman
Justin Thurman

Reputation: 3

C character array printing numbers insted of the assigned charaters

I am working on an assignment to solve a maze created from a 2d character array. To test the program I made a simple 4x4 maze. But the maze, when printed to the screen is comprised of numbers. I am very confused at how this even happens. Any help would be appreciated.

The assignment is this: 

char *maze[4][4];
for (int i=0; i < 4; ++i)
{
    maze[0][i] = "#";
    maze[3][i] = "#";
    maze[1][i] = ".";
}
maze[2][0] = "#";
maze[2][3] = "#";
maze[2][1] = ".";
maze[2][2] = ".";

and printing is here: 

for(int i =0; i < 4; ++i)
{
    for(int j = 0; j < 4; ++j)
    {
        printf("%c",maze[i][j]);
    }
    printf("\n");
}

I expected it to print this:

####
....
#..#
####

But instead it prints:

0000
2222
0220
0000

Upvotes: 0

Views: 120

Answers (3)

mooiamaduck
mooiamaduck

Reputation: 2156

There's a type mis-match in the printf call. You're printing maze[i][j], which is a char * (string), as a character (%c). I suggest turning on compiler warnings to catch these kinds of errors; gcc found the issue when I tried to compile your code.

The reason it prints a number instead of a character is because printf is interpreting the address of the string maze[i][j] as an ASCII code point and printing the corresponding character. For you compiler, the addresses of "#" and "." happen to result in the characters 0 and 2 being printed. It was different in my case; when I compiled your code, the program printed EOT and ACK.

The nicest solution would be to declare maze as an array of chars instead of strings.

char maze[4][4] = {
  { '#', '#', '#', '#' },
  { '.', '.', '.', '.' },
  { '#', '.', '.', '#' },
  { '#', '#', '#', '#' }
};

Upvotes: 1

ouah
ouah

Reputation: 145919

maze[0][i] = "#";

should be

maze[0][i] = '#';

and char *maze[4][4]; should be char maze[4][4];

"#" is a string literal, use '#' to have a character constant.

If you really want to use string literals of one character you have to use the %s conversion specification instead of %c in your original program.

Upvotes: 6

bottaio
bottaio

Reputation: 5093

The problem lies in this line

printf("%c",maze[i][j]);

you are not using characters, but char pointers (aka strings) so this should go like:

printf("%s",maze[i][j]);

Or you can also use characters instead as described in other answer.

Upvotes: 5

Related Questions