double size array initialize ?

What's the maximum expected size?    

If it's less than a million or so, it's often simpler to just declare it to be that size.   That's assuming this is running on some reasonably modern computer with many megabytes of memory.

For larger sizes, use malloc:

double * Array;

Array = malloc( sizeof( double ) * TheActualSize );

\
....   use   Array here as usual.

free( Array );  // when you're done using the array.

expected size is not too big.

actually i want to learn its sense, logic.

i do it like above,

double *p;

p=(*double) malloc(sizeof(double) * what ? )

what is the actual size ?

and also, how can i reach the integer value @ point[1][2], if it is 8*8 array. like *(p+i) ? i is counter.

Check out the wrapper functions in the following links:
http://code.axter.com/allocate2darray.h
http://code.axter.com/allocate2darray.c

thanks deepudeepam , it is really good .

last thing about the line; " double *p[] " to understand,

double's is in the code because numbers can be large, right ? instead of double may i use int ?
and why we use *p[], not only *p ?

thanks so much.

Here's example for creating a buffer for a 2-D array:

void **Allocate2DArray(int TypeSize, int x, int y)
{
      void **ppi            = malloc(x*sizeof(void*));
      void *pool            = malloc(x*y*TypeSize);
      unsigned char *curPtr = pool;
      int i;
      if (!ppi || !pool)
      {  /* Quit if either allocation failed */
            if (ppi) free(ppi);
            if (pool) free(pool);
            return NULL;
      }

      for(i = 0; i < x; i++)
      {
            *(ppi + i) = curPtr;
            curPtr += y*TypeSize;
      }
      return ppi;
}



Example usage:
int **My2DChar = Allocate2DArray(sizeof(int), x, y);

ok.. that's all thanks..