Regarding double and triple pointers/double dimension arrays

Question

So, i was playing with C pointers and pointer arithmetic since i'm not entirely comfortable with them. I came up with this code.

char* a[5] = { "Hi", "My", "Name", "Is" , "Dennis"};
char** aPtr = a; // This is acceptable because 'a' is double pointer
char*** aPtr2 = &aPtr; // This is also acceptable because they are triple pointers
//char ***aPtr2 = &a // This is not acceptable according to gcc 4.8.3, why ?

//This is the rest of the code, the side notes are only for checking
printf("%s\n",a[0]); //Prints Hi
printf("%s\n",a[1]); //Prints My
printf("%s\n",a[2]); //Prints Name
printf("%s\n",a[3]); //Prints Is
printf("%s\n",a[4]); //Prints Dennis

printf("%s\n",*(a+0)); //Prints Hi
printf("%s\n",*(a+1)); //Prints My
printf("%s\n",*(a+2)); //Prints Name
printf("%s\n",*(a+3)); //Prints Is
printf("%s\n",*(a+4)); //Prints Dennis

printf("%s\n",*(*(aPtr2) +0)); //Prints Hi 
printf("%s\n",*(*(aPtr2) +1)); //Prints My // ap = a, *ap = *a, *(ap)+1 = *a+1 ?
printf("%s\n",*(*(aPtr2) +2)); //Prints Name
printf("%s\n",*(*(aPtr2) +3)); //Prints Is
printf("%s\n",*(*(aPtr2) +4)); //Prints Dennis

char*** aPtr2 = &a is not acceptable according to gcc 4.8.3, why?

Sorry forgot to add compiler warning:

warning: initialization from incompatible pointer type [enabled by default]

It maybe unclear what I'm trying to say, so I had to add this links:

• This is the code that works: http://ideone.com/4ePj4h. (line 7. commented out)
• This is the code that does not work: http://ideone.com/KMG7OS. (line 6. commented out)

Notice the commented out lines.

Answer 1

a is the address of a buffer of 5 ptrs and is immutable (i.e. its a fixed ptr). If you allowed

char ***aPtr2 = &a;

then

*aPtr2 = &a[3];

would actually modify the address a (which is verboten).