paradoxical subreg problem

[Richard Kenner]

    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.

    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.

    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.