Code:
2.2: 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 2.4.) 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. The pointer is to be known by the name "p," and can point to any char (or contiguous array of chars) anywhere.
As usual, a picture is worth a thousand words. The statements
char a[] = "hello";
char *p = "world";
would result in 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 the example above, both a[3] and p[3] happen to be the character 'l', but the compiler gets there differently. (See also questions 17.19 and 17.20.)
2.6: Why doesn't sizeof properly report the size of an array which is a parameter to a function?
The sizeof operator reports the size of the pointer parameter which the function actually receives (see question 2.4).
Basically, you're taking the sizeof a pointer, rather than the array that it points to.