CSE360_HW6

CSE360_HW6 - $1551 $15 ANSWER3 : A-STAK: .skip (512) *( 4)...

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Abdulwahab Alkharasahi & Hany Rashwan CSE 360 HW#6 ANSWRWE 1: Instr A B CCR PC M[$E000} # BYTES #CYCLES Initial Value $B3 $2B $09 $E000 TBA $2B $2B $01 $E001 $17 1 2 CLRB $B3 $00 $04 $E001 $5F 1 2 ABA $DE $2B $08 $E001 $1B 1 2 BNE SZERO $B3 $2B $09 $E00A $26 2 3 LDAA $02 $2B $01 $E002 $86 2 2 STAA $1004 $B3 $2B $09 $E003 $B7 3 4 BEQ FIN $B3 $2B $09 $E002 $27 2 3 TAP $B3 $2B $B3 $E001 $06 1 2 ANSWER2: SP A B CCR IX M[00B6] $00B7 $B3 $4D $0A $1551 $7F PSHX $00B5 $B3 $4D $0A $1551 $15 PULA $00B6 $15 $4D $0A $1551 $15 PULB $00B7 $15 51D $0A
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Unformatted text preview: $1551 $15 ANSWER3 : A-STAK: .skip (512) *( 4) BSTAK: B-set.BSTAK, %sp C-set x, %r id [%r], %r sub %sp, 4% sp st %r1, [%sp] D-id[%sp], %r2 add %sp, 4, %sp set x, %r1 st%r2, [%r1] ANSWER 4: A-Ordering from most to least terms of flexibility: 1. Dispatcher 2. Vectored linking 3. Hard addressing Since the dispatcher has more routines, it will be more flexible B-Ordering fastest to slowest terms of speed: 1-Hard addressing 2-Vectored linking 3-Dispatcher 4-Slowest Hard addressing is the only way that has to accesses memory at first....
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CSE360_HW6 - $1551 $15 ANSWER3 : A-STAK: .skip (512) *( 4)...

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