Ich bin mir nicht sicher, aber folgendes Besipiel kann ich nicht mit UUs alleine lösen (geht das überhaupt):
-------------------------------------
if (a==0)
S1
else
if (b==0) && (c==0)
S2
else
S3
endif
endif
-------------------------------------
Zur Erinnerung:
pred (a),px<cond>,py<cond> (P)
a..........Test Condition zb. (a>7)
P..........Predicate Bit
px<cond>...Verändere px, mit der Tabelle unten
a P px | UU iUU | PU iPU | OR iOR | AND iAND |
----------+----------+----------+----------+----------+
x 0 x | 0 0 | x x | x x | x x |
0 1 x | 0 1 | 0 1 | x 1 | 0 x |
1 1 x | 1 0 | 1 0 | 1 x | x 0 |
Lösung mit iOR:
pred (a==0), p0<UU>,p1<iUU>
pred (b==0), p1<UU>,p2<iUU> (p1)
pred (c==0), p1<UU>,p2<iOR> (p1)
S1 (p0)
S2 (p1)
S3 (p2)
| (init) | step 1 | step 2 | step 3 |
a b c | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 |
-------+----------+----------+----------+----------+
0 0 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 0 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 1 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 1 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
1 0 0 | x x x | 0 1 x | 0 1 0 | 0 1 0 |
1 0 1 | x x x | 0 1 x | 0 1 0 | 0 0 1 |
1 1 0 | x x x | 0 1 x | 0 0 1 | 0 0 1 |
1 1 1 | x x x | 0 1 x | 0 0 1 | 0 0 1 |
Andere Lösung mit iPU:
pred (a==0), p0<UU>,p1<iUU>
pred (b==0), p1<UU>,p2<iUU> (p1)
pred (c==0), p1<UU>,p2<iPU> (p1)
S1 (p0)
S2 (p1)
S3 (p2)
| (init) | step 1 | step 2 | step 3 |
a b c | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 |
-------+----------+----------+----------+----------+
0 0 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 0 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 1 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 1 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
1 0 0 | x x x | 0 1 x | 0 1 0 | 0 1 0 |
1 0 1 | x x x | 0 1 x | 0 1 0 | 0 0 1 |
1 1 0 | x x x | 0 1 x | 0 0 1 | 0 0 1 |
1 1 1 | x x x | 0 1 x | 0 0 1 | 0 0 1 |
Keine Lösung nur PUs:
pred (a==0), p0<PU>,p1<iPU>
pred (b==0), p1<PU>,p2<iPU> (p1)
pred (c==0), p1<PU>,p2<iPU> (p1)
S1 (p0)
S2 (p1)
S3 (p2)
| (init) | step 1 | step 2 | step 3 |
a b c | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 |
-------+----------+----------+----------+----------+
0 0 0 | x x x | 1 0 x | 1 0 x | 1 0 x |
0 0 1 | x x x | 1 0 x | 1 0 x | 1 0 x |
0 1 0 | x x x | 1 0 x | 1 0 x | 1 0 x |
0 1 1 | x x x | 1 0 x | 1 0 x | 1 0 x |
1 0 0 | x x x | 0 1 x | 0 1 0 | 0 1 0 |
1 0 1 | x x x | 0 1 x | 0 1 0 | 0 0 1 |
1 1 0 | x x x | 0 1 x | 0 0 1 | 0 0 1 |
1 1 1 | x x x | 0 1 x | 0 0 1 | 0 0 1 |
wenn a=0 (obere 4 zeilen) wird p1 gleich 0 gesetzt.
da aber PUs ich schreibt wenn das predicate false ist wird p2 nie
geschrieben.
Lösung nur PUs (eine extra Instruction):
pred (a==0), p0<PU>,p1<iPU>
pred (a==0), p0<PU>,p2<iPU>
pred (b==0), p1<PU>,p2<iPU> (p1)
pred (c==0), p1<PU>,p2<iPU> (p1)
S1 (p0)
S2 (p1)
S3 (p2)
| (init) | step 1 | step 2 | step 3 | step 4 |
a b c | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 |
-------+----------+----------+----------+----------+----------+
0 0 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 | 1 0 0 |
0 0 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 | 1 0 0 |
0 1 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 | 1 0 0 |
0 1 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 | 1 0 0 |
1 0 0 | x x x | 0 1 x | 0 1 1 | 0 1 0 | 0 1 0 |
1 0 1 | x x x | 0 1 x | 0 1 1 | 0 1 0 | 0 0 1 |
1 1 0 | x x x | 0 1 x | 0 1 1 | 0 0 1 | 0 0 1 |
1 1 1 | x x x | 0 1 x | 0 1 1 | 0 0 1 | 0 0 1 |
Schrit 2 inititalisiert p2, daher passt alles.
Keine Lösung nur UUs:
pred (a==0), p0<UU>,p1<iUU>
pred (b==0), p1<UU>,p2<iUU> (p1)
pred (c==0), p1<UU>,p2<iUU> (p1)
S1 (p0)
S2 (p1)
S3 (p2)
| (init) | step 1 | step 2 | step 3 |
a b c | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 | p0 p1 p2 |
-------+----------+----------+----------+----------+
0 0 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 0 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 1 0 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
0 1 1 | x x x | 1 0 x | 1 0 0 | 1 0 0 |
1 0 0 | x x x | 0 1 x | 0 1 0 | 0 1 0 |
1 0 1 | x x x | 0 1 x | 0 1 0 | 0 0 1 |
1 1 0 | x x x | 0 1 x | 0 0 1 | 0 0 0 |
1 1 1 | x x x | 0 1 x | 0 0 1 | 0 0 0 |
Im Schritt 3 wenn a=b=1 (die letzten beiden Zeilen) ist p1=0. UU
schreibt in in diesem Fall unabhängig vom alten ergebnis des vergleichs
eine 0
Fällt euch eine Lösung nur mi UUs ein?Bez. der 24 Bit ISA wäre es wichtig nur eine Predicate-Condition zu haben, ansonsten gehen uns die Bits aus. Die einzige Variante die übrig bliebe wäre nur mit PUs zu arbeiten.
Oder fällt irgend wem von euch eine Lösung mit UUs ein. g Martin
