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Published bySidney Hutcheon Modified about 1 year ago

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מבנה המחשב – מבוא למחשבים ספרתיים The ISA of the Simplified DLX

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Problem #1 LEGEND: r16 = f r17 = g r18 = h r19 = i r20 = j r21 = k xorr1r19r20 beqzr11 addr16r17r18 subr16r16r19 if(i==j) goto L1; f=g+h; L1: f=f-i;

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Problem #2 xorr1r19r20 bnezr12 addr16r17r18 beqzr11 subr16r17r18 if(i==j) skip to else f=g+h; else f=g-h;

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Problem #3 Address of A[0] = Astart addir4r0Astart addr1r4r19 lwr2r10 addr17r17r2 addr19r19r20 xorr3r19r18 bnezr3-6 LOOP: g=g + A[i] ; i=i+j; if((i!=h) goto LOOP;

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Problem #4 Address of save[0] = Sstart addir4r0Sstart addr1r4r19 subr2r20r19 addr3r4r2 lwr5r10 lwr6r30 xorr7r5r6 bnezr72 addir19r191 beqzr0-9 while ( save[i] == save[k-i] ) i=i+1;

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Problem #5 addir4r210 bnezr42 addr16r19r20 beqzr011 addir4r4-1 bnezr42 addr16r17r18 beqzr07 addir4r4-1 bnezr42 subr16r17r18 beqzr03 addir4r4-1 bnezr41 subr16r19r20 switch(k){ case 0: f=i+j; break; case 1: f=g+h; break; case 2: f=g-h; break; case 3: f=i-j; }

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