Pointer to Pointer with argv

Question

Based on my understanding of pointer to pointer to an array of characters,

% ./pointer one two

argv           
+----+          +----+
| .  |   --->   | .         


        

Answer 1

Because argv is a pointer to pointer to char, it follows that argv[1] is a pointer to char. The printf() format %s expects a pointer to char argument and prints the null-terminated array of characters that the argument points to. Since argv[1] is not a null pointer, there is no problem.

(*argv)[1] is also valid C, but (*argv) is equivalent to argv[0] and is a pointer to char, so (*argv)[1] is the second character of argv[0], which is / in your example.

Answer 2

Indexing a pointer as an array implicitly dereferences it. p[0] is *p, p[1] is *(p + 1), etc.

Answer 3

It's more convenient to think of [] as an operator for pointers rather than arrays; it's used with both, but since arrays decay to pointers array indexing still makes sense if it's looked at this way. So essentially it offsets, then dereferences, a pointer.

So with argv[1], what you've really got is *(argv + 1) expressed with more convenient syntax. This gives you the second char * in the block of memory pointed at by argv, since char * is the type argv points to, and [1] offsets argv by sizeof(char *) bytes then dereferences the result.

(*argv)[1] would dereference argv first with * to get the first pointer to char, then offset that by 1 * sizeof(char) bytes, then dereferences that to get a char. This gives the second character in the first string of the group of strings pointed at by argv, which is obviously not the same thing as argv[1].

So think of an indexed array variable as a pointer being operated on by an "offset then dereference a pointer" operator.