paradoxical subreg problem

[Richard Earnshaw]

> 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.
> 

I was under the impression that we would only create paradoxical subregs 
of a mem if LOAD_EXTEND_OP returns either ZERO_EXTEND or SIGN_EXTEND for 
the mode of the mem (ie, we won't create such paradoxical subregs if 
LOAD_EXTEND_OP returns NIL).  That is, the operation is an alternate way 
of expressing the extend operation, because that is what will happen in 
practice.

Given the above, then on such machines

	(subreg:SI (mem:QI X) 0)

is defined for all 32 bits.

R.