What is the difference between char *a and char a[]?

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 crucial 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';