Out of ISO/IEC 9899-1990:
6.3.6 Additive Operators
Semantics
[...] If the pointer operand points to an element of an array object, and
the array is large enough, the result points to an element offset from the
original element such that the difference of the subscripts of the
resulting and original array elements equals the integral expression.
So if p + i == q, then q - p == i. If in addition "i < 0", the q must be
less then p. The only constraint on i is that it is an integral expression,
and the constraints on p and q are that they point into the same "array
object." Hell, ptrdiff_t (a signed integral type) may have fewer bits
then a pointer, and the algebra must certainly work there.
I'm pretty sure this applies to your example by noting the section:
6.3.2.1 Array Subscripting
Semantics
[...] The definition of the subscript operator [] is that E1[E2] is
identical to (*(E1+(E2))). Because if the conversion rules that apply
to the binary + operator, if E1 is an array object (equivalently, a
pointer to the initial element of an array object) and E2 is an integer,
E1[E2] designates the E2-th element of E1 (counting from zero).
So I would say that any compiler that generates code that prints your
message "C does not promote..." is broken, no matter how many bits in a
pointer.
-- Steve Williams "The woods are lovely, dark and deep. steve@icarus.com But I have promises to keep, steve@picturel.com and lines to code before I sleep, http://www.picturel.com And lines to code before I sleep."
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