There is a lot of difference!


char a[] = "string";
char *a = "string";


The declaration char a[] asks for space for 7 characters and see that its known by
the name "a". In contrast, the declaration char *a, asks for a place that holds a
pointer, to be known by the name "a". This pointer "a" can point anywhere. In this
case its pointing to an anonymous array of 7 characters, which does have any name
in particular. Its just present in memory with a pointer keeping track of its location.




char a[] = "string";

   +----+----+----+----+----+----+------+
a: | s  |  t |  r |  i |  n |  g | '\0' |
   +----+----+----+----+----+----+------+
   a[0] a[1] a[2] a[3] a[4] a[5] a[6]


char *a = "string";

+-----+          +---+---+---+---+---+---+------+
|  a: | *======> | s | t | r | i | n | g | '\0' |
+-----+          +---+---+---+---+---+---+------+
Pointer          Anonymous array



It is curcial to know that a[3] generates different code depending on whether a is
an array or a pointer. When the compiler sees the expression a[3] and if a is an
array, it starts at the location "a", goes three elements past it, and returns the
character there. When it sees the expression a[3] and if a is a pointer, it starts
at the location "a", gets the pointer value there, adds 3 to the pointer value, and
gets the character pointed to by that value. 

If a is an array, a[3] is three places past a. If a is a pointer, then a[3] is three
places past the memory location pointed to by a. In the example above, both a[3]
and a[3] return the same character, but the way they do it is different! 

Doing something like this would be illegal.


char *p = "hello, world!";
p[0] = 'H';