Thread: Dynamic char array allocation and assignment

I'm allocating memory for a 8X8 char array:

Code:

char **game;
game = (char**)malloc(8 * sizeof(char*));
for(i = 0; i < 8; i++)   
   game[i] = (char*)malloc(8 * sizeof(char));

then try to assign a character from a char array to each column in the first row game[0][0] = r, game[0][1] = b and so on...

Code:

char myArray[8] = "rbnqknbr";
i=0;
for(int j=0; j<8; j++)   
   game[i][j] = myArray[j];

Code compiles with no errors but the cmd line crashes when I run the program. Any thoughts?

You should be checking return value from malloc ; other than that, more of code might be revealing.

Can you explain this a bit more? check the return value of malloc?
Thats my code, at least thats where I'm stuck with my exercise.

this:

Code:

game[0][0] = 'r';
game[0][1] = 'b';     
game[0][2] = 'n';
game[0][3] = 'q';     
game[0][4] = 'k';     
game[0][5] = 'n';     
game[0][6] = 'b';     
game[0][7] = 'r';

has the same result btw.

malloc - C++ Reference Read the Return Value entry.

If that is all of your code, then it runs fine for me. With the exception of declaring j inside the for loop, no problems. Paste your whole program in.

So to explain what's going on in your program (and if I say it wrong, I'm sure someone will slap my hand) :

C doesn't have a true pass by reference. That is, game in your function is not some alias to game in main. It is a local variable, a pointer to pointer that expects an address as argument for the function.
When you call allocate_memory(game), game is evaluated for it's value (an address to a char pointer) which is passed to the function and stored in the function's variable game. So is game pointing to a char pointer when you call the function?

All that said, the simple solution is to not pass any values, but to return a pointer to pointer.

Thanks for your reply and explanation.
Do you mean that I should return a pointer to pointer inside allocate_memory() function?
How do I implement that?

all my code:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

void allocate_memory(char **game)
{
    game = (char**)malloc(8 * sizeof(char*));
    int i;
    for(i = 0; i < 8; i++)
       game[i] = (char*)malloc(8 * sizeof(char));      
}

void main ()
{
     char **game;
     char pawns[8] = "rbnqknbr";

     allocate_memory(game);
     for(int j=0; j<8; j++)
         game[0][j] = pawns[j];
}

Code:

char ** allocate_memory( void )
{
char **ptr = {...} ;
...
return ptr ;
}

Change the prototype of the function

Or alternatively, use an additional level of pointers to emulate pass by reference :

Looks complex, but you're basically just replacing every instance of game in allocate_memory with *game. The only tricky part is you need (*game)[i] since *game[i] means something different (specifically, take game[i] as a pointer and dereference it, which is different from what you want - take game as a pointer, dereference it to get an array, and then get entry i from that array).

No real reason to pick this versus the "return a pointer" approach until you have a function which allocates more than one pointer. Since you can only return one value, you'd need a different approach in that case.

Code:

void allocate_memory(char ***game)
{
   *game = malloc(8 * sizeof(char*));
    int i;
    for(i = 0; i < 8; i++)
       (*game)[i] = malloc(8 * sizeof(char));     
}
 
int main ()
{
     char **game;
     char pawns[8] = "rbnqknbr";
 
     allocate_memory(&game);
     for(int j=0; j<8; j++)
         game[0][j] = pawns[j];
}