77 lines
2.3 KiB
NASM
77 lines
2.3 KiB
NASM
mov 0,ct0 ; num
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mov 0,ct1 ; vdp2 address
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mov 0,ct3 ; output
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mvi 12345,mc0 ; m0[0] (12345)
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;; output: m3
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base10_loop:
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mov 0,ct0
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mvi div10_unsigned,pc
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;; [X ] [Y ] [D1 ]
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mov mc0,y mov 1,lop ; mvi imm,pc delay slot (executed twice)
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;; after function return:
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mov all,mc0 ; m0[1] (1234)
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mov all,pl ; ??? why can't this happen in the next instruction?
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;; mod10: multiply by 10:
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sl mov alu,a
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sl mov alu,a
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sl mov alu,a
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add mov alu,a
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add mov alu,a mov 0,ct0 ; restore ct0 after 'mov all,mc1'
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;; a: 12340
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mov mc0,a mov all,pl
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;; a: 12345 m0[0]
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;; p: 12340
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;; mod10: subtract (a - p)
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sub mov alu,a
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;; a: 5
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;; store digit in m3
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mov mc0,p clr a mov all,mc3 ; m0[1]
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;; p: 1234
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;; a: 0
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add mov alu,a mov 0,ct0
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;; a: 1234
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jmp nz,base10_loop
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mov all,mc0 ; jmp delay slot
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endi
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nop
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;; argument: ry
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;; return: a ← ry / 10
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;; maximum ry is somewhere between 2 ^ 21 and 2 ^ 22
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div10_unsigned:
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;; 1 / 10 * (2 ^ 24) ~= 1677722 = 0x19999a
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mvi 1677722,rx
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mov mul,p clr a
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ad2 mov alu,a
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;; ALH: [nmlkjihgfedcba9876543210________]
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;; ALL: [76543210________________________]
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clr a mov alh,pl ; alh is (a >> 16)
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add mov alu,a
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;; ALL: [nmlkjihgfedcba9876543210________]
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;; rotate left 24 requires fewer cycles than shift right 8
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rl8 mov alu,a
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rl8 mov alu,a
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rl8 mov alu,a
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;; ALL: [________nmlkjihgfedcba9876543210]
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;; mask 24 bit result
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mvi 0xffffff,pl
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;; return to caller ; reset ct0
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btm
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and mov alu,a mov 0,ct0
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