paradoxical subreg problem

[law@redhat.com]

In message <10201281935.AA25716@vlsi1.ultra.nyu.edu>, Richard Kenner writes:
 >     Don't assume you can break it into two expressions.  Consider the
 >     expression as it stands (and as combine creates it).
 > 
 > Sure, but I'm trying to define what it means by comparison with
 > two expressions.
 > 
 >     So with your assertions in mind are these two expresions equivalent?
 > 
 >     (and:SI (subreg:SI (mem:QI) 0) (const_int 255))
 > 
 >     (subreg:SI (mem:QI X) 0)

Are you sure?

Are the bits outside the mode of SUBREG_REG undefined or "don't care"?  That
is the crux of the issue.

If those bits are undefined, then those expressions are not equivalent as
the result of the first expression has 24 zeros in its high bits whereas
the second expression has 24 undefined bits.  If those bits are "don't care"
then the compiler can treat those expressions as equivalent.

jeff