paradoxical subreg problem

[law@redhat.com]

In message <10201281902.AA25419@vlsi1.ultra.nyu.edu>, Richard Kenner writes:
 >     I'll start simple.  On a big endian machine, can this expression be
 >     optimized into true/false at compile time, or must it be run-time
 >     computed?
 > 
 >     (eq (subreg:SI (mem/s:QI (plus:SI (reg:SI 3 %r3)
 >                     (const_int 15 [0xf])) 1) 0)
 >         (mem/s:SI (plus:SI (reg:SI 3 %r3)
 >               (const_int 12 [0xc])) 1))
 > 
 > This is equivalent to:
 > 
 >     (set (reg:QI xx) (mem/s:QI (plus:SI (reg:SI 3 %r3)
 > 				        (const_int 15 [0xf])) 0))
 >     (eq (subreg:SI (reg:QI xx) 0)
 >         (mem/s:SI (plus:SI (reg:SI 3 %r3)
 > 	                   (const_int 12 [0xc])) 1))
 > 
 > 
 > on all machines.
Don't assume you can break it into two expressions.  Consider the
expression as it stands (and as combine creates it).

 >     According to my reading, the compiler is allowed to optimize the
 >     expression into (true) because the bits outside of QImode on the
 >     subreg are "don't care bits" -- meaning they can have any value that
 >     is convenient to us.
 > 
 >     Agree/Disagree?
 > 
 > I agree.
OK.

 > 
 >     Now consider if byte loads zero extend.  Does your answer change?  In
 >     the subreg arm, those "don't care" bits, have a well defined meaning --
 >     ie, we can't pretend they have whatever value is convenient for us.
 >     So, unless we have some more specific knowledge about the other arm,
 >     then this expression must be evaluated at runtime.
 > 
 > I disagree.  We "know" what they will be, but the undefined semantics still
 > holds.  So this can also be true.
OK.

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)


jeff