"abcdef"[3] == 3["abcdef"], but why?

[Chris Torek]

In article <781 at cc.helsinki.fi> TEITTINEN at cc.helsinki.fi writes:
>Could someone explain to me what a C compiler does when it runs into
>expression 3["abcdef"]? 

3["abcdef"] is of the form `e1 [ e2 ]' (e1 and e2 are arbitrary
expressions).  Every C compiler converts this internally to

	*( (e1) + (e2) )

(actually, compilers need only work *as if* they had done this
conversion, although many really do it).  So we really have

	*( 3 + "abcdef" )

To find out what this means, if anything, first try evaluating the
expression `3+"abcdef"'.  This is

	<value: int: 3> + <object: array 7 of char: `abcdef\0'>

Apply the rule for array objects in rvalue contexts: change `array N
of T' to `pointer to T', whose value is the address of the first (0th)
element of the array:

	<value: int: 3> + <value: char *: points to `a' in `abcdef\0'>

We now have an expression of the form <int>+<pointer>, so we move to
the int'th element of the array that starts at *<pointer>.  Here we
move to the 3rd element of `abcdef\0', which is the letter `c'.  Now
we have

	<value: char *: points to `c' in `abcdef\0'>

Now we can put the indirection back:

	*( <value: char *: points to `c' in `abcdef\0'> )

The operand for `*' is a pointer, so we get the object to which the
pointer points:

	<object: char: `c'>

Since e1[e2] is equivalent to *(e1+e2), if we reverse the expressions,
the only change is in the order of the operands to `+'.  The result of
<pointer>+<int> is the same as that of <int>+<pointer>, so we wind up
with <object: char: `c'> again.
-- 
In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7163)
Domain:	chris at mimsy.umd.edu	Path:	uunet!mimsy!chris