Question 6.2

But I heard that char a[] was identical to char *a.

Not at all. (What you heard has to do with formal parameters to functions; see question 6.4.) Arrays are not pointers. The array declaration char a[6] requests that space for six characters be set aside, to be known by the name ``a.'' That is, there is a location named ``a'' at which six characters can sit. The pointer declaration char *p, on the other hand, requests a place which holds a pointer, to be known by the name ``p.'' This pointer can point almost anywhere: to any char, or to any contiguous array of chars, or nowhere (see also questions 5.1 and 1.30).

As usual, a picture is worth a thousand words. The declarations

	char a[] = "hello";
	char *p = "world";

would initialize data structures which could be represented like this:

	   +---+---+---+---+---+---+
	a: | h | e | l | l | o |\0 |
	   +---+---+---+---+---+---+
	   +-----+     +---+---+---+---+---+---+
	p: |  *======> | w | o | r | l | d |\0 |
	   +-----+     +---+---+---+---+---+---+

It is important to realize that a reference like x[3] generates different code depending on whether x is an array or a pointer. Given the declarations above, when the compiler sees the expression a[3], it emits code to start at the location ``a,'' move three past it, and fetch the character there. When it sees the expression p[3], it emits code to start at the location ``p,'' fetch the pointer value there, add three to the pointer, and finally fetch the character pointed to. In other words, a[3] is three places past (the start of) the object named a, while p[3] is three places past the object pointed to by p. In the example above, both a[3] and p[3] happen to be the character 'l', but the compiler gets there differently.

References: K&R2 Sec. 5.5 p. 104
CT&P Sec. 4.5 pp. 64-5

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