filling a 2d array random

Question

this code will good for n<20, but for n=40 give me access violation error: this code will fill X and O random.

 int i=0,j=0;
 int x=0,y=0;
 int n=40;
 for(i=0;i<n;i++)
 {
 for(j=0;j<n;j++)
 arr[i][j]='O';
 }
 srand(clock());
 for(i=0;i<n*n;i++)
 {
 x = rand()%n;
 y = rand()%n;
 if(arr[x][y] == 'O') arr[x][y]='X';
 }

Declare:

 arr = (char**)malloc(n);
 for(i=0;i<n;i++)
 arr[i] = (char*)malloc(n);

• CanSpice: Why not just randomly set the value to X or O in the first loop?
• Drahakar: Can we see your arr declaration too?
• Michael Dorgan: Man I hate 2D array malloc. I never get it right. My guess is that there's nothing that forces malloc above to keep memory contiguous so it will randomly fail.
• KCH: you can use memory from stack if the array will be small, it's a lot easier.

Answer 1

change

arr = (char**)malloc(n);

to

arr = (char**)malloc(n*sizeof(char*));

Answer 2

you could do :-

 for(i=0;i<n;i++)
 {
 for(j=0;j<n;j++)
 arr[i][j]= ((rand() % 2) == 0) ? 'O' : 'X';
 }

and make sure your array is n by n. Instead of those multiple mallocs, which will allocate memory from all over the place...

 arr = (char**)malloc( n * n * sizeof(char));

Answer 3

for(i=0;i<n*n;i++)
{
 x = rand()%n;
 y = rand()%n;
 if(arr[x][y] == 'O') arr[x][y]='X';
 ...

n*n? arr only has n elements and arr[0...n-1] each only have n elements. If x or y is >= n, you'll be accessing elements past the end of your array and causing undefined behaviour. In this case your lucky because it causes an access violation.

That, and arr = (char**)malloc(n); should be arr = (char**)malloc(n * sizeof(char*));.

Answer 4

If n is constant, or your compiler supports C99, you can initalize it as a 1-d, then cast it to a pointer-to array:

int i;
char (*arr)[n] = malloc(n*n);
char *vals = (char*)arr;
for(i=0; i<n*n; i++)
 vals[i] = (rand()%2)?'X':'O';
//now, use arr[y][x] as usual