101 lines
2.5 KiB
NASM
101 lines
2.5 KiB
NASM
prog START 0
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. beri stevke do newline (\n = 10), nato evalviraj
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LDX #1 . zacnemo z 1 za lazje preverjanje pogoja v loop_e
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loop RD device . FA.dev
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COMP #10 . newline
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JEQ evaluate
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SUB #48 . ASCII koda za 0
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+STCH stevke, X
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TIX #-1 . primerjava TIX tu nima pomena
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J loop
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evaluate . T zacasno hrani rezultat, na koncu v A
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LDT #0
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. S hrani potence st. deset
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LDS #1
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. zmanjsaj X za 1
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RMO X, A
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SUB #1
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RMO A, X
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. nalozi stevko, pomnozi s S, shrani oz. pristej rez. v T
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loop_e LDCH stevke, X
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MULR S, A
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ADDR A, T
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. povecaj potenco
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RMO S, A
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MUL #10
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RMO A, S
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. zmanjsaj X
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RMO X, A
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SUB #1
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RMO A, X
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COMP #0
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JGT loop_e
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. ce je rezultat nic koncaj sicer izracunaj fakulteto in preberi naslednjo vrstico
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RMO T, A
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COMP #0
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JEQ HALT
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JSUB stackinit
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JSUB fakulteta_r
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. JSUB izpis
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STA temp
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CLEAR A
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CLEAR T
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CLEAR S
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CLEAR X
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J prog
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HALT J HALT
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END prog
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fakulteta_r +STL @stackptr . link reg. na sklad
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JSUB stackpush
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+STA @stackptr . parameter na sklad
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JSUB stackpush
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SUB #1 . pripravimo podproblem
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COMP #1 . zaustavitveni pogoj
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JEQ exit . ce je z. p. izpolnjen RSUB, v A je vrednost podproblema
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JSUB fakulteta_r
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exit JSUB stackpop
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RMO A, B . premakni vrednost podroblema v B
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+LDA @stackptr . nalozi parameter klicujoce funkcije ("starša")
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MULR B, A . pomnozi, shrani v A
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JSUB stackpop
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+LDL @stackptr
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RSUB
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stackinit STA oldvalue
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CLEAR A
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LDA #stack
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STA stackptr
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CLEAR A
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LDA oldvalue
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RSUB
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stackpush STA oldvalue
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CLEAR A
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LDA stackptr
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ADD #3
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STA stackptr
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CLEAR A
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LDA oldvalue
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RSUB
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stackpop STA oldvalue
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CLEAR A
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LDA stackptr
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SUB #3
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STA stackptr
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CLEAR A
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LDA oldvalue
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RSUB
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temp WORD 0
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device BYTE X'FA'
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stevke BYTE 100
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oldvalue WORD 0
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stackptr WORD 0
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stack RESW 1000
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