From: bouncer@dev.null   
      
   kiran wrote:   
      
   > Here is a code snippet:   
   > =============   
   > int **ptr;   
   > int intArr[3][5];   
   > ptr = intArr;   
   > =============   
   >   
   > This gives the error on g++ compiler:   
   > error: cannot convert ¡int [3][5]¢ to ¡int**¢ in assignment   
   >   
   > I have tried other variants as well, like   
   > int const **ptr = intArr;   
   > and   
   > ptr = &intArr[0];   
   >   
   > I still got the same error. Can anyone explain the reason here.   
      
   The declaration   
      
     int intArr[3][5];   
      
   declares intArr to be an array of 3 arrays of 5 ints.  The language's   
   decay rule says that the name of this array can be implictly converted   
   to a pointer to its first element.  Thus, we can write:   
      
     int (*p1)[5] = intArr;   
      
   Note that p1 is not a pointer to a pointer to an int; it is a pointer to   
   an array of 5 ints, pointing to intArr[0].   
      
   In contrast, if we write:   
      
     int **p2;   
      
   We're saying that p2 points to a pointer to an int.  If it were legal to   
   write   
      
     int **p2 = intArr;   
      
   then where is the pointer that p2 points to?  There is none, which is   
   why this conversion has to be illegal.   
      
   HTH,   
      
   - Wil   
      
      
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