On 12/18/23 22:32, Peter Maydell wrote:
+ if (s->nv && s->nv2 && ri->nv2_redirect_offset) {
Again, s->nv test is redundant.
+ /*
+ * Some registers always redirect to memory; some only do so if
+ * HCR_EL2.NV1 is 0, and some only if NV1 is 1 (these come in
+ * pairs which share an offset; see the table in R_CSRPQ).
+ */
+ if (ri->nv2_redirect_offset & NV2_REDIR_NV1) {
+ nv2_mem_redirect = s->nv1;
+ } else if (ri->nv2_redirect_offset & NV2_REDIR_NO_NV1) {
+ nv2_mem_redirect = !s->nv1;
+ } else {
+ nv2_mem_redirect = true;
+ }
I wondered if it would be clearer with the "both" case having both bits set. While I see
that the first defined offset is 0x20, offset 0x00 is still reserved and *could* be used.
At which point ri->nv2_redirect_offset would need a non-zero value for a zero offset.
Maybe clearer as
nv2_mem_redirect = (ri->nv2_redirect_offset &
(s->nv1 ? NV2_REDIR_NV1_1 : NV2_REDIR_NV1_0));
?
This is more verbose for the (common?) case of redirect regardless of nv1, so
maybe not.
+ if (s->nv2_mem_be) {
+ mop |= MO_BE;
+ }
MO_BSWAP is host dependent -- needs
mop |= (s->nv2_mem_be ? MO_BE : MO_LE);
r~
