(unless (= high 0)
(inst ldah reg high reg)))))
((or (unsigned-byte 32) (signed-byte 64) (unsigned-byte 64))
+ ;; Since it took NJF and CSR a good deal of puzzling to work out
+ ;; (a) what a previous version of this was doing and (b) why it
+ ;; was wrong:
+ ;;
+ ;; write VALUE = a_63 * 2^63 + a_48-62 * 2^48
+ ;; + a_47 * 2^47 + a_32-46 * 2^32
+ ;; + a_31 * 2^31 + a_16-30 * 2^16
+ ;; + a_15 * 2^15 + a_0-14
+ ;;
+ ;; then, because of the wonders of sign-extension and
+ ;; twos-complement arithmetic modulo 2^64, if a_15 is set, LDA
+ ;; (which sign-extends its argument) will add
+ ;;
+ ;; (a_15 * 2^15 + a_0-14 - 65536).
+ ;;
+ ;; So we need to add that 65536 back on, which is what this
+ ;; LOGBITP business is doing. The same applies for bits 31 and
+ ;; 47 (bit 63 is taken care of by the fact that all of this
+ ;; arithmetic is mod 2^64 anyway), but we have to be careful that
+ ;; we consider the altered value, not the original value.
+ ;;
+ ;; I think, anyway. -- CSR, 2003-09-26
(let* ((value1 (if (logbitp 15 value) (+ value (ash 1 16)) value))
- (value2 (if (logbitp 31 value) (+ value (ash 1 32)) value1))
- (value3 (if (logbitp 47 value) (+ value (ash 1 48)) value2)))
+ (value2 (if (logbitp 31 value1) (+ value1 (ash 1 32)) value1))
+ (value3 (if (logbitp 47 value2) (+ value2 (ash 1 48)) value2)))
(inst lda reg (ldb (byte 16 32) value2) zero-tn)
+ ;; FIXME: Don't yet understand these conditionals. If I'm
+ ;; right, surely we can just consider the zeroness of the
+ ;; particular bitfield, not the zeroness of the whole thing?
+ ;; -- CSR, 2003-09-26
(unless (= value3 0)
(inst ldah reg (ldb (byte 16 48) value3) reg))
(unless (and (= value2 0) (= value3 0))