Fix LRA subreg calculation for big-endian targets
LRA was using a subreg offset of 0 whenever constraints matched two operands with different modes. That leads to an invalid offset (and ICE) on big-endian targets if one of the modes is narrower than a word. E.g. if a (reg:SI X) is matched to a (reg:QI Y), the big-endian subreg should be (subreg:QI (reg:SI X) 3) rather than (subreg:QI (reg:SI X) 0). But this raises the issue of what the behaviour should be when the matched operands occupy different numbers of registers. Should the register numbers match, or should the locations of the lsbs match? Although the documentation isn't clear, reload went for the second interpretation (which seems the most natural to me): /* On a REG_WORDS_BIG_ENDIAN machine, point to the last register of a multiple hard register group of scalar integer registers, so that for example (reg:DI 0) and (reg:SI 1) will be considered the same register. */ So I think this means that we can/must use the lowpart offset unconditionally, rather than trying to separate out the multi-register case. This also matches the LRA handling of constant integers, which already uses lowpart subregs. The patch fixes gcc.target/aarch64/sve/extract_[34].c for aarch64_be. 2018-01-30 Richard Sandiford <richard.sandiford@linaro.org> gcc/ * lra-constraints.c (match_reload): Use subreg_lowpart_offset rather than 0 when creating partial subregs. From-SVN: r257177
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