Execute Computer/Zero: Difference between revisions
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=={{header|ALGOL 68}}==
<
# instructions #
INT nop = 0, lda = 1, sta = 2, add = 3, sub = 4, brz = 5, jmp = 6, stp = 7;
Line 96:
execute( "1+255", ( LDA 3, ADD 4, STP 0, 1, 255 ) )
END
</syntaxhighlight>
{{out}}
<pre>
Line 109:
</pre>
=={{header|
==={{header|Applesoft BASIC}}===
<syntaxhighlight lang="gwbasic"> 0 FOR P = 1 TO 5: READ S$: PRINT S$" : ";: FOR MEM = 0 TO 31: ON LEN (S$) > 0 GOSUB 90:A = VAL (S$): GOSUB 2: NEXT MEM: GOSUB 7: PRINT A: NEXT P: END
1 LET A = V: RETURN
2 LET P$ = MID$ (P$,1,MEM) + CHR$ (A) + MID$ (P$,MEM + 2,31 - MEM): PRINT MID$ ("",1 ^ FRE (0));: RETURN
Line 133 ⟶ 134:
300 DATA "FIBONACCI",46,79,109,78,47,77,48,145,171,80,192,46,224,1,1,0,8,1,
400 DATA "LINKED LIST",45,111,69,112,71,0,78,0,171,79,192,46,224,32,0,28,1,0,0,0,6,0,2,26,5,20,3,30,1,22,4,24
500 DATA "PRISONER",0,0,224,0,0,35,157,178,35,93,174,33,127,65,194,32,127,64,192,35,93,33,126,99,</
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
The original version of the virtual machine has a provision of entering user choices in the Prisoner's dilema game. It works allowing the user to enter a number when executing a Stop instruction. There is no simple way to do this with Rosetta Code instructions that the program should return a value when it reaches a Stop instruction. As a result, all the versions displayed in the Rosetta Code examples, simply returns a zero when it reaches the stop instruction.
<syntaxhighlight lang="Delphi">
{structure that holds name of program and code}
type TProgram = record
Name, Code: string;
end;
{List of programs}
var ProgList: array [0..4] of TProgram =(
(Name: 'Two plus two'; Code: 'LDA 3 ADD 4 STP 2 2'),
(Name: 'Seven times eight'; Code: 'LDA 12 ADD 10 STA 12 LDA 11 SUB 13 STA 11 BRZ 8 JMP 0 LDA 12 STP 8 7 0 1'),
(Name: 'The Fibonacci sequence'; Code: 'LDA 14 STA 15 ADD 13 STA 14 LDA 15 STA 13 LDA 16 SUB 17 BRZ 11 STA 16 JMP 0 LDA 14 STP 1 1 0 8 1'),
(Name: 'Linked list'; Code: 'LDA 13 ADD 15 STA 5 ADD 16 STA 7 NOP STA 14 NOP BRZ 11 STA 15 JMP 0 LDA 14 STP LDA 0 0 28 1 0 0 0 6 0 2 26 5 20 3 30 1 22 4 24'),
(Name: 'Prisoner’s Dilemma'; Code: '0 0 STP NOP LDA 3 SUB 29 BRZ 18 LDA 3 STA 29 BRZ 14 LDA 1 ADD 31 STA 1 JMP 2 LDA 0 ADD 31 STA 0 JMP 2 LDA 3 STA 29 LDA 0 ADD 30 ADD 3 STA 0 LDA 1 ADD 30 ADD 3 STA 1 JMP 2 0 1 3')
);
{Defines opcode function call}
type TOpProc = function(Opcode: byte): boolean;
{Defines argument type for opcode}
type TArgType = (atNone,atAddress,atMemory);
{Defines Opcode information}
type TOpcode = record
OpCode: integer;
ArgType: TArgType;
Mnemonic: string[3];
Proc: TOpProc;
end;
{Simulated memory, program counter and accumulator}
var Memory: array [0..31] of byte;
var ProgramCounter: integer;
var Accumulator: byte;
{Routines to carry out actions of opcodes}
function DoNoOp(Operand: byte): boolean;
{ ($00) = NOP - (no operation)}
begin
Inc(ProgramCounter);
Result:=True;
end;
function DoLDA(Operand: byte): boolean;
{ ($20) LDA - (load accumulator) - a := memory [xxxxx]}
begin
Accumulator:=Memory[Operand];
Inc(ProgramCounter);
Result:=True;
end;
function DoSTA(Operand: byte): boolean;
{ ($40) STA - (store accumulator) - memory [xxxxx] := a}
begin
Memory[Operand]:=Accumulator;
Inc(ProgramCounter);
Result:=True;
end;
function DoADD(Operand: byte): boolean;
{ ($60) ADD - (add) - a := a + memory [xxxxx]}
begin
Accumulator:=Accumulator + Memory[Operand];
Inc(ProgramCounter);
Result:=True;
end;
function DoSUB(Operand: byte): boolean;
{ ($80) SUB - (subtract) - a := a – memory [xxxxx]}
begin
Accumulator:=Accumulator - Memory[Operand];
Inc(ProgramCounter);
Result:=True;
end;
function DoBRZ(Operand: byte): boolean;
{ ($A0) BRZ - (branch on zero) - if a = 0 then goto xxxxx}
begin
if Accumulator=0 then ProgramCounter:=Operand
else Inc(ProgramCounter);
Result:=True;
end;
function DoJMP(Operand: byte): boolean;
{ ($C0) JMP - (jump) - goto xxxxx}
begin
ProgramCounter:=Operand;
Result:=True;
end;
function DoSTP(Operand: byte): boolean;
{ ($E0) STP - (stop)}
begin
Result:=False;
end;
{Table of opcodes}
var OpTable: array [0..7] of TOpcode= (
(Opcode: $00; ArgType: atNone; Mnemonic: 'NOP'; Proc: DoNoOp),
(Opcode: $20; ArgType: atMemory; Mnemonic: 'LDA'; Proc: DoLDA),
(Opcode: $40; ArgType: atMemory; Mnemonic: 'STA'; Proc: DoSTA),
(Opcode: $60; ArgType: atMemory; Mnemonic: 'ADD'; Proc: DoADD),
(Opcode: $80; ArgType: atMemory; Mnemonic: 'SUB'; Proc: DoSUB),
(Opcode: $A0; ArgType: atAddress; Mnemonic: 'BRZ'; Proc: DoBRZ),
(Opcode: $C0; ArgType: atAddress; Mnemonic: 'JMP'; Proc: DoJMP),
(Opcode: $E0; ArgType: atNone; Mnemonic: 'STP'; Proc: DoSTP));
procedure ClearMemory;
{Set all memory locations to zero}
var I: integer;
begin
for I:=0 to High(Memory) do
Memory[I]:=0;
end;
function DecodeMnemonic(NMem: string): TOpcode;
{Convert 3-char Mnemonic to opcode info}
var I: integer;
begin
for I:=0 to High(OpTable) do
if NMem=OpTable[I].Mnemonic then
begin
Result:=OpTable[I];
exit;
end;
raise Exception.Create('Opcode Not Found');
end;
procedure AssembleProgram(Prog: string);
{Convert text source code into machine code and store in memory}
var Inx,MemInx,IntAtom,T: integer;
var Atom: string;
var OC: TOpcode;
begin
ClearMemory;
MemInx:=0;
Inx:=1;
while true do
begin
{Get one space delimited atom}
Atom:=ExtractToken(Prog,' ',Inx);
{exit if end of source code}
if Atom='' then break;
{Is the atom text (Mnemonic) or number?}
if Atom[1] in ['A'..'Z','a'..'z'] then
begin
{Convert Mnemonic to opcode info}
OC:=DecodeMnemonic(Atom);
{Get machine value of opcode}
IntAtom:=OC.OpCode;
{Is operand of of memory or address type?}
if OC.ArgType in [atMemory,atAddress] then
begin
{if so combine opcode and operand}
Atom:=ExtractToken(Prog,' ',Inx);
T:=StrToInt(Atom);
IntAtom:=IntAtom or T;
end;
end
else IntAtom:=StrToInt(Atom);
{Store machine code in memory}
Memory[MemInx]:=IntAtom;
{Point to next memory location}
Inc(MemInx);
end;
end;
function DecodeOpcode(Inst: byte; var Opcode: TOpcode): boolean;
{Convert machine instruction to opcode information}
var I: integer;
var Op: byte;
begin
Result:=True;
{Get opcode part of instruction}
Op:=$E0 and Inst;
{Look for it in table}
for I:=0 to High(OpTable) do
if OpTable[I].OpCode=Op then
begin
Opcode:=OpTable[I];
exit;
end;
Result:=False;
end;
function ExecuteInstruction(Instruct: byte): boolean;
{Execute a single instruction}
var I: integer;
var Opcode: TOpcode;
var Operand: byte;
begin
if not DecodeOpcode(Instruct,Opcode) then raise Exception.Create('Illegal Opcode');
{Separate instruction and operand}
Operand:=$1F and Instruct;
{Execute instruction}
Result:=Opcode.Proc(Operand);
end;
procedure DisplayInstruction(Memo: TMemo; PC: integer);
{Display instruction information}
var Opcode: TOpcode;
var Inst,Operand: integer;
var S: string;
begin
Inst:=Memory[PC];
if not DecodeOpcode(Inst,Opcode) then raise Exception.Create('Illegal Opcode');
Operand:=Inst and $1F;
S:=IntToStr(PC)+' $'+IntToHex(Inst,2);
S:=S+' '+Opcode.Mnemonic;
case Opcode.ArgType of
atMemory:
begin
S:=S+' Mem['+IntToStr(Operand)+']=';
S:=S+' '+IntToStr(Memory[Operand]);
end;
atAddress: S:=S+' '+IntToStr(Operand);
end;
Memo.Lines.Add(S);
end;
procedure ExecuteProgram(Memo: TMemo; Prog: TProgram);
{Executed program until stop instruction reached}
begin
Memo.Lines.Add(Prog.Name);
AssembleProgram(Prog.Code);
ProgramCounter:=0;
Accumulator:=0;
repeat
begin
DisplayInstruction(Memo,ProgramCounter);
end
until not ExecuteInstruction(Memory[ProgramCounter]);
Memo.Lines.Add('Result='+IntToStr(Accumulator));
Memo.Lines.Add('');
end;
procedure ShowComputerZero(Memo: TMemo);
{Execute and display all five sample programs.}
begin
ExecuteProgram(Memo,ProgList[0]);
ExecuteProgram(Memo,ProgList[1]);
ExecuteProgram(Memo,ProgList[2]);
ExecuteProgram(Memo,ProgList[3]);
ExecuteProgram(Memo,ProgList[4]);
end;
</syntaxhighlight>
{{out}}
<pre>
Two plus two
0 $23 LDA Mem[3]= 2
1 $64 ADD Mem[4]= 2
2 $E0 STP
Result=4
Seven times eight
0 $2C LDA Mem[12]= 0
1 $6A ADD Mem[10]= 8
2 $4C STA Mem[12]= 0
3 $2B LDA Mem[11]= 7
4 $8D SUB Mem[13]= 1
5 $4B STA Mem[11]= 7
6 $A8 BRZ 8
7 $C0 JMP 0
0 $2C LDA Mem[12]= 8
1 $6A ADD Mem[10]= 8
2 $4C STA Mem[12]= 8
3 $2B LDA Mem[11]= 6
4 $8D SUB Mem[13]= 1
5 $4B STA Mem[11]= 6
6 $A8 BRZ 8
7 $C0 JMP 0
0 $2C LDA Mem[12]= 16
1 $6A ADD Mem[10]= 8
2 $4C STA Mem[12]= 16
3 $2B LDA Mem[11]= 5
4 $8D SUB Mem[13]= 1
5 $4B STA Mem[11]= 5
6 $A8 BRZ 8
7 $C0 JMP 0
0 $2C LDA Mem[12]= 24
1 $6A ADD Mem[10]= 8
2 $4C STA Mem[12]= 24
3 $2B LDA Mem[11]= 4
4 $8D SUB Mem[13]= 1
5 $4B STA Mem[11]= 4
6 $A8 BRZ 8
7 $C0 JMP 0
0 $2C LDA Mem[12]= 32
1 $6A ADD Mem[10]= 8
2 $4C STA Mem[12]= 32
3 $2B LDA Mem[11]= 3
4 $8D SUB Mem[13]= 1
5 $4B STA Mem[11]= 3
6 $A8 BRZ 8
7 $C0 JMP 0
0 $2C LDA Mem[12]= 40
1 $6A ADD Mem[10]= 8
2 $4C STA Mem[12]= 40
3 $2B LDA Mem[11]= 2
4 $8D SUB Mem[13]= 1
5 $4B STA Mem[11]= 2
6 $A8 BRZ 8
7 $C0 JMP 0
0 $2C LDA Mem[12]= 48
1 $6A ADD Mem[10]= 8
2 $4C STA Mem[12]= 48
3 $2B LDA Mem[11]= 1
4 $8D SUB Mem[13]= 1
5 $4B STA Mem[11]= 1
6 $A8 BRZ 8
8 $2C LDA Mem[12]= 56
9 $E0 STP
Result=56
The Fibonacci sequence
0 $2E LDA Mem[14]= 1
1 $4F STA Mem[15]= 0
2 $6D ADD Mem[13]= 1
3 $4E STA Mem[14]= 1
4 $2F LDA Mem[15]= 1
5 $4D STA Mem[13]= 1
6 $30 LDA Mem[16]= 8
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
9 $50 STA Mem[16]= 8
10 $C0 JMP 0
0 $2E LDA Mem[14]= 2
1 $4F STA Mem[15]= 1
2 $6D ADD Mem[13]= 1
3 $4E STA Mem[14]= 2
4 $2F LDA Mem[15]= 2
5 $4D STA Mem[13]= 1
6 $30 LDA Mem[16]= 7
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
9 $50 STA Mem[16]= 7
10 $C0 JMP 0
0 $2E LDA Mem[14]= 3
1 $4F STA Mem[15]= 2
2 $6D ADD Mem[13]= 2
3 $4E STA Mem[14]= 3
4 $2F LDA Mem[15]= 3
5 $4D STA Mem[13]= 2
6 $30 LDA Mem[16]= 6
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
9 $50 STA Mem[16]= 6
10 $C0 JMP 0
0 $2E LDA Mem[14]= 5
1 $4F STA Mem[15]= 3
2 $6D ADD Mem[13]= 3
3 $4E STA Mem[14]= 5
4 $2F LDA Mem[15]= 5
5 $4D STA Mem[13]= 3
6 $30 LDA Mem[16]= 5
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
9 $50 STA Mem[16]= 5
10 $C0 JMP 0
0 $2E LDA Mem[14]= 8
1 $4F STA Mem[15]= 5
2 $6D ADD Mem[13]= 5
3 $4E STA Mem[14]= 8
4 $2F LDA Mem[15]= 8
5 $4D STA Mem[13]= 5
6 $30 LDA Mem[16]= 4
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
9 $50 STA Mem[16]= 4
10 $C0 JMP 0
0 $2E LDA Mem[14]= 13
1 $4F STA Mem[15]= 8
2 $6D ADD Mem[13]= 8
3 $4E STA Mem[14]= 13
4 $2F LDA Mem[15]= 13
5 $4D STA Mem[13]= 8
6 $30 LDA Mem[16]= 3
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
9 $50 STA Mem[16]= 3
10 $C0 JMP 0
0 $2E LDA Mem[14]= 21
1 $4F STA Mem[15]= 13
2 $6D ADD Mem[13]= 13
3 $4E STA Mem[14]= 21
4 $2F LDA Mem[15]= 21
5 $4D STA Mem[13]= 13
6 $30 LDA Mem[16]= 2
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
9 $50 STA Mem[16]= 2
10 $C0 JMP 0
0 $2E LDA Mem[14]= 34
1 $4F STA Mem[15]= 21
2 $6D ADD Mem[13]= 21
3 $4E STA Mem[14]= 34
4 $2F LDA Mem[15]= 34
5 $4D STA Mem[13]= 21
6 $30 LDA Mem[16]= 1
7 $91 SUB Mem[17]= 1
8 $AB BRZ 11
11 $2E LDA Mem[14]= 55
12 $E0 STP
Result=55
Linked list
0 $2D LDA Mem[13]= 32
1 $6F ADD Mem[15]= 28
2 $45 STA Mem[5]= 0
3 $70 ADD Mem[16]= 1
4 $47 STA Mem[7]= 0
5 $3C LDA Mem[28]= 1
6 $4E STA Mem[14]= 0
7 $3D LDA Mem[29]= 22
8 $AB BRZ 11
9 $4F STA Mem[15]= 28
10 $C0 JMP 0
0 $2D LDA Mem[13]= 32
1 $6F ADD Mem[15]= 22
2 $45 STA Mem[5]= 60
3 $70 ADD Mem[16]= 1
4 $47 STA Mem[7]= 61
5 $36 LDA Mem[22]= 2
6 $4E STA Mem[14]= 1
7 $37 LDA Mem[23]= 26
8 $AB BRZ 11
9 $4F STA Mem[15]= 22
10 $C0 JMP 0
0 $2D LDA Mem[13]= 32
1 $6F ADD Mem[15]= 26
2 $45 STA Mem[5]= 54
3 $70 ADD Mem[16]= 1
4 $47 STA Mem[7]= 55
5 $3A LDA Mem[26]= 3
6 $4E STA Mem[14]= 2
7 $3B LDA Mem[27]= 30
8 $AB BRZ 11
9 $4F STA Mem[15]= 26
10 $C0 JMP 0
0 $2D LDA Mem[13]= 32
1 $6F ADD Mem[15]= 30
2 $45 STA Mem[5]= 58
3 $70 ADD Mem[16]= 1
4 $47 STA Mem[7]= 59
5 $3E LDA Mem[30]= 4
6 $4E STA Mem[14]= 3
7 $3F LDA Mem[31]= 24
8 $AB BRZ 11
9 $4F STA Mem[15]= 30
10 $C0 JMP 0
0 $2D LDA Mem[13]= 32
1 $6F ADD Mem[15]= 24
2 $45 STA Mem[5]= 62
3 $70 ADD Mem[16]= 1
4 $47 STA Mem[7]= 63
5 $38 LDA Mem[24]= 5
6 $4E STA Mem[14]= 4
7 $39 LDA Mem[25]= 20
8 $AB BRZ 11
9 $4F STA Mem[15]= 24
10 $C0 JMP 0
0 $2D LDA Mem[13]= 32
1 $6F ADD Mem[15]= 20
2 $45 STA Mem[5]= 56
3 $70 ADD Mem[16]= 1
4 $47 STA Mem[7]= 57
5 $34 LDA Mem[20]= 6
6 $4E STA Mem[14]= 5
7 $35 LDA Mem[21]= 0
8 $AB BRZ 11
11 $2E LDA Mem[14]= 6
12 $E0 STP
Result=6
Prisoner’s Dilemma
0 $00 NOP
1 $00 NOP
2 $E0 STP
Result=0
Elapsed Time: 611.666 ms.
</pre>
=={{header|EasyLang}}==
<syntaxhighlight>
proc run name$ mem[] . .
write name$ & ": "
pc = 1
len mem[] 32
repeat
ppc = mem[pc]
op = ppc div 32
addr = ppc mod 32 + 1
pc += 1
if op = 1
acc = mem[addr]
elif op = 2
mem[addr] = acc
elif op = 3
acc = (acc + mem[addr]) mod 255
elif op = 4
acc = (acc - mem[addr]) mod 255
elif op = 5
if acc = 0
pc = addr
.
elif op = 6
pc = addr
.
until op = 7 or pc > 32
.
print acc
.
run "2+2" [ 35 100 224 2 2 ]
run "7*8" [ 44 106 76 43 141 75 168 192 44 224 8 7 0 1 ]
run "Fibonacci" [ 46 79 109 78 47 77 48 145 171 80 192 46 224 1 1 0 8 1 ]
run "List" [ 45 111 69 112 71 0 78 0 171 79 192 46 224 32 0 28 1 0 0 0 6 0 2 26 5 20 3 30 1 22 4 24 ]
run "Prisoner" [ 0 0 224 0 0 35 157 178 35 93 174 33 127 65 194 32 127 64 192 35 93 33 126 99 ]
</syntaxhighlight>
{{out}}
<pre>
2+2: 4
7*8: 56
Fibonacci: 55
List: 6
Prisoner: 0
</pre>
=={{header|Forth}}==
{{works with|gforth|0.7.3}}
<br>
<
\ Execute Computer/Zero
Line 332 ⟶ 892:
bye
</syntaxhighlight>
{{out}}
Line 556 ⟶ 1,116:
0 1 : NOP 0 -> 0 2
0 2 : STP 0 -> 0 -1
=={{header|Go}}==
<syntaxhighlight lang="go">package main
import (
"bufio"
"fmt"
"io"
"regexp"
"strconv"
"strings"
)
const (
NOP = iota
LDA
STA
ADD
SUB
BRZ
JMP
STP
)
var opcodes = map[string]int{
"NOP": NOP,
"LDA": LDA,
"STA": STA,
"ADD": ADD,
"SUB": SUB,
"BRZ": BRZ,
"JMP": JMP,
"STP": STP,
}
var reIns = regexp.MustCompile(
`\s*(?:(\w+):)?\s*` + // label
`(NOP|LDA|STA|ADD|SUB|BRZ|JMP|STP)?\s*` + // opcode
`(\w+)?\s*` + // argument
`(?:;([\w\s]+))?`) // comment
type ByteCode [32]int
type instruction struct {
Label string
Opcode string
Arg string
}
type Program struct {
Instructions []instruction
Labels map[string]int
}
func newInstruction(line string) (*instruction, error) {
match := reIns.FindStringSubmatch(line)
if match == nil {
return nil, fmt.Errorf("syntax error: '%s'", line)
}
return &instruction{Label: match[1], Opcode: match[2], Arg: match[3]}, nil
}
func Parse(asm io.Reader) (*Program, error) {
var p Program
p.Labels = make(map[string]int, 32)
scanner := bufio.NewScanner(asm)
lineNumber := 0
for scanner.Scan() {
if instruction, err := newInstruction(scanner.Text()); err != nil {
return &p, err
} else {
if instruction.Label != "" {
p.Labels[instruction.Label] = lineNumber
}
p.Instructions = append(p.Instructions, *instruction)
lineNumber++
}
}
if err := scanner.Err(); err != nil {
return nil, err
}
return &p, nil
}
func (p *Program) Compile() (ByteCode, error) {
var bytecode ByteCode
var arg int
for i, ins := range p.Instructions {
if ins.Arg == "" {
arg = 0
} else if addr, err := strconv.Atoi(ins.Arg); err == nil {
arg = addr
} else if addr, ok := p.Labels[ins.Arg]; ok {
arg = addr
} else {
return bytecode, fmt.Errorf("unknown label %v", ins.Arg)
}
if opcode, ok := opcodes[ins.Opcode]; ok {
bytecode[i] = opcode<<5 | arg
} else {
bytecode[i] = arg
}
}
return bytecode, nil
}
func floorMod(a, b int) int {
return ((a % b) + b) % b
}
func Run(bytecode ByteCode) (int, error) {
acc := 0
pc := 0
mem := bytecode
var op int
var arg int
loop:
for pc < 32 {
op = mem[pc] >> 5
arg = mem[pc] & 31
pc++
switch op {
case NOP:
continue
case LDA:
acc = mem[arg]
case STA:
mem[arg] = acc
case ADD:
acc = floorMod(acc+mem[arg], 256)
case SUB:
acc = floorMod(acc-mem[arg], 256)
case BRZ:
if acc == 0 {
pc = arg
}
case JMP:
pc = arg
case STP:
break loop
default:
return acc, fmt.Errorf("runtime error: %v %v", op, arg)
}
}
return acc, nil
}
func Execute(asm string) (int, error) {
program, err := Parse(strings.NewReader(asm))
if err != nil {
return 0, fmt.Errorf("assembly error: %v", err)
}
bytecode, err := program.Compile()
if err != nil {
return 0, fmt.Errorf("compilation error: %v", err)
}
result, err := Run(bytecode)
if err != nil {
return 0, err
}
return result, nil
}
func main() {
examples := []string{
`LDA x
ADD y ; accumulator = x + y
STP
x: 2
y: 2`,
`loop: LDA prodt
ADD x
STA prodt
LDA y
SUB one
STA y
BRZ done
JMP loop
done: LDA prodt ; to display it
STP
x: 8
y: 7
prodt: 0
one: 1`,
`loop: LDA n
STA temp
ADD m
STA n
LDA temp
STA m
LDA count
SUB one
BRZ done
STA count
JMP loop
done: LDA n ; to display it
STP
m: 1
n: 1
temp: 0
count: 8 ; valid range: 1-11
one: 1`,
`start: LDA load
ADD car ; head of list
STA ldcar
ADD one
STA ldcdr ; next CONS cell
ldcar: NOP
STA value
ldcdr: NOP
BRZ done ; 0 stands for NIL
STA car
JMP start
done: LDA value ; CAR of last CONS
STP
load: LDA 0
value: 0
car: 28
one: 1
; order of CONS cells
; in memory
; does not matter
6
0 ; 0 stands for NIL
2 ; (CADR ls)
26 ; (CDDR ls) -- etc.
5
20
3
30
1 ; value of (CAR ls)
22 ; points to (CDR ls)
4
24`,
`LDA 3
SUB 4
STP 0
0
255`,
`LDA 3
SUB 4
STP 0
0
1`,
`LDA 3
ADD 4
STP 0
1
255`,
}
for _, asm := range examples {
if result, err := Execute(asm); err == nil {
fmt.Println(result)
} else {
fmt.Println(err)
}
}
}
</syntaxhighlight>
{{out}}
<pre>
4
56
55
6
1
255
0
</pre>
=={{header|J}}==
Line 563 ⟶ 1,405:
Here's one approach which might be sufficiently close to the task requirements:
<
assemble1=: {{
Line 607 ⟶ 1,449:
while. 0<:pc do. exec1'' end.
acc
}}</
With this implementation, we can assemble and run representations of the five suggested programs:
<
4
exec assemble 'LDA 12; ADD 10; STA 12; LDA 11; SUB 13; STA 11; BRZ 8; JMP; LDA 12; STP; 8; 7; 0; 1'
Line 620 ⟶ 1,462:
6
exec assemble 'NOP; NOP; STP; NOP; LDA 3; SUB 29; BRZ 18; LDA 3; STA 29; BRZ 14; LDA 1; ADD 31; STA 1; JMP 2; LDA; ADD 31; STA; JMP 2; LDA 3; STA 29; LDA 1; ADD 30; ADD 3; STA 1; LDA; ADD 30; ADD 3; STA; JMP 2; 0; 1; 3'
0</
=== Alternate approach ===
Line 626 ⟶ 1,468:
Another approach would be to eliminate the 'assemble' command and implement the opcodes as native J commands (which build a representation of the memory space). To conform with J syntax, the right arguments of these opcodes is required (so you need a 0 even for NOP and STP). The implementation of <code>exec</code> remains much the same as before, but it's convenient to have exec save the code to memory. In other words:
<
(m+{.y),}.y
:
Line 654 ⟶ 1,496:
while. 0<:pc do. exec1'' end.
acc
}}</
and:
<
4
exec LDA 12 ADD 10 STA 12 LDA 11 SUB 13 STA 11 BRZ 8 JMP 0 LDA 12 STP 0 8 7 0 1
Line 667 ⟶ 1,509:
6
exec NOP 0 NOP 0 STP 0 NOP 0 LDA 3 SUB 29 BRZ 18 LDA 3 STA 29 BRZ 14 LDA 1 ADD 31 STA 1 JMP 2 LDA 0 ADD 31 STA 0 JMP 2 LDA 3 STA 29 LDA 1 ADD 30 ADD 3 STA 1 LDA 0 ADD 30 ADD 3 STA 0 JMP 2 0 1 3
0</
=={{header|Java}}==
Line 673 ⟶ 1,515:
{{works with|Java|8}}
<
import static java.lang.Math.min;
import static java.util.stream.Collectors.toMap;
Line 978 ⟶ 1,820:
}
}
</syntaxhighlight>
{{out}}
Line 989 ⟶ 1,831:
1
255
0
</pre>
=={{header|jq}}==
'''Adapted from [[#Wren|Wren]]'''
{{works with|jq}}
'''Also works with gojq, the Go implementation of jq'''
This entry assumes all the programs have been placed in a text file
in which each program takes the form of a header followed
by the instructions, one per line, followed by at least one blank line.
The header is assumed to have the form:
<pre>
; title
</pre>
<syntaxhighlight lang="jq">
### Utility
def trim: sub("^ +";"") | sub(" +$";"");
### Computer/Zero
def NOP: 0;
def LDA: 1;
def STA: 2;
def ADD: 3;
def SUB: 4;
def BRZ: 5;
def JMP: 6;
def STP: 7;
def ops: {"NOP": NOP, "LDA": LDA, "STA": STA, "ADD": ADD,
"SUB": SUB, "BRZ": BRZ, "JMP": JMP, "STP": STP};
# Input: the program in the form of an array of strings,
# each string corresponding to an input line of the form
# "INSTR N" or "N"
# Output: an array of integers
def load:
map([splits(" *")] as $split
| $split[0] as $instr
| (if ($split|length == 2) then $split[1]|tonumber
else 0
end) as $addr
| if ops[$instr]
then ops[$instr] * 32 + $addr
else try ($instr|tonumber) catch 0
end );
# input: an array as produced by `load`
def interp:
{ acc: 0, pc: 0, mem: .}
| until(.break;
(.mem[.pc] % 32) as $addr
| ((.mem[.pc] - $addr) / 32) as $instr
| .pc += 1
| if $instr == LDA then .acc = .mem[$addr]
elif $instr == STA then .mem[$addr] = .acc
elif $instr == ADD then .acc += .mem[$addr]
| if .acc > 255 then .acc += -256 else . end
elif $instr == SUB then .acc += (- .mem[$addr])
| if .acc < 0 then .acc += 256 else . end
elif $instr == BRZ
then if .acc == 0 then .pc = $addr else . end
elif $instr == JMP then .pc = $addr
else .
end
| .break = $instr == STP or .pc > 31 )
| .acc;
# Assume the input file consists of several programs, each structured as:
# ; program name
# one instruction per line
#
def task:
def init: map_values(null);
foreach (inputs, null) as $line ({};
if $line == null then .program = .buffer
elif $line[0:1] == ";" then init | .title = $line
else ($line|trim) as $line
| if $line == "" then .program = .buffer | .buffer = []
else .buffer += [$line]
end
end)
| .title as $title
| .program
| if length == 0 then empty
else
$title, (load|interp)
end ;
task
</syntaxhighlight>
'''Program file - scrollable window''' (programs.txt)
<div style="overflow:scroll; height:400px;">
<pre>
; 2+2
LDA 3
ADD 4
STP
2
2
; 7 * 8
LDA 12
ADD 10
STA 12
LDA 11
SUB 13
STA 11
BRZ 8
JMP 0
LDA 12
STP
8
7
0
1
; fibonacci
LDA 14
STA 15
ADD 13
STA 14
LDA 15
STA 13
LDA 16
SUB 17
BRZ 11
STA 16
JMP 0
LDA 14
STP
1
1
0
8
1
; linkedList
LDA 13
ADD 15
STA 5
ADD 16
STA 7
NOP
STA 14
NOP
BRZ 11
STA 15
JMP 0
LDA 14
STP
LDA 0
0
28
1
0
0
0
6
0
2
26
5
20
3
30
1
22
4
24
; prisoner
0
0
STP
NOP
LDA 3
SUB 29
BRZ 18
LDA 3
STA 29
BRZ 14
LDA 1
ADD 31
STA 1
JMP 2
LDA 0
ADD 31
STA 0
JMP 2
LDA 3
STA 29
LDA 0
ADD 30
ADD 3
STA 0
LDA 1
ADD 30
ADD 3
STA 1
JMP 2
0
1
3
</pre>
</div>
{{output}}
'''Invocation:''' jq -nR -f execute-computer-zero.jq programs.txt
<pre>
"; 2+2"
4
"; 7 * 8"
56
"; fibonacci"
55
"; linkedList"
6
"; prisoner"
0
</pre>
=={{header|Julia}}==
<
ip::Int
ram::Vector{UInt8}
Line 1,168 ⟶ 2,231:
run(compile(t))
end
</
<pre>
Compiled 6 lines.
Line 1,187 ⟶ 2,250:
=== Interpreter version with labels ===
Uses the Python examples.
<
ip, accum, isready, ram = 0x1, 0x0, true, zeros(UInt8, 32)
NOP() = (ip = mod1(ip + 1, 32))
Line 1,481 ⟶ 2,544:
interpret(t)
end
</
<pre>
Program completed with accumulator value 4.
Line 1,513 ⟶ 2,576:
===Execute (2+2, 7*8)===
Per primary task, just execute first two examples:
<
pc = 0,
acc = 0,
Line 1,542 ⟶ 2,605:
-- object code derived from js sim code
print("2 + 2 = " .. vmz:boot():load({[0]=35,100,224,2,2}):exec().acc)
print("7 x 8 = " .. vmz:boot():load({[0]=44,106,76,43,141,75,168,192,44,224,8,7,0,1}):exec().acc)</
{{out}}
<pre>2 + 2 = 4
Line 1,548 ⟶ 2,611:
===Disassembler (7*8)===
Assembly syntax doesn't seem well-specified (except as shown in descriptive text, which is presumably just pseudocode), but next step might be to add a minimal disassembler:
<
local mnem, imax = { [0]="NOP", "LDA", "STA", "ADD", "SUB", "BRZ", "JMP", "STP" }
for pc = 31,0,-1 do imax=pc if self.mem[pc]~=0 then break end end
Line 1,562 ⟶ 2,625:
print(vmz:dism())
print("Disassembly of 7 x 8 (POST-RUN):")
print(vmz:exec():dism())</
{{out}}
<pre>Disassembly of 7 x 8 (PRE-RUN):
Line 1,597 ⟶ 2,660:
===Assembler (fibo, list)===
Add an assembler (same minimal syntax as disassembler) for the next two examples:
<
vmz.assm = function(self, source)
local function oa2b(oper, addr) return oper * 32 + addr end
Line 1,617 ⟶ 2,680:
NOP 5 NOP 20 NOP 3 NOP 30 NOP 1 NOP 22 NOP 4 NOP 24
]]
print("Linked list = " .. vmz:boot():assm(srclist):exec().acc)</
{{out}}
<pre>Fibonacci = 55
Line 1,623 ⟶ 2,686:
===Prisoner===
Some additional helper routines, then multiple rounds:
<
vmz.poke = function(self,addr,byte) self.mem[addr]=byte end
vmz.entr = function(self,byte) self.mem[self.pc]=byte self.pc=self.pc+1 end
Line 1,639 ⟶ 2,702:
vmz:exec()
print("Round: " .. r .. " Move: " .. mvnm[move] .. " Player: " .. vmz:peek(0) .. " Computer: " .. vmz:peek(1))
end</
{{out}}
<pre>Prisoner (priming) = 0
Line 1,647 ⟶ 2,710:
Round: 4 Move: coop Player: 6 Computer: 6
Round: 5 Move: defe Player: 6 Computer: 9</pre>
=={{header|Nim}}==
We don’t use an assembly language but define the memory values using functions named NOP, LDA, STA, etc. To be able to use Uniform Function Call Syntax (i.e. call without parentheses),
all these functions have an argument, even NOP and STP.
In the last example, “Prisoners”, an input is expected from the user. We defined a function “enterValue” which stores the input value into the byte at address referenced by the PC then increments the PC by one. We have chosen the same player strategy as the one used by the Lua program and, of course, we get the same result.
<syntaxhighlight lang="Nim">import std/strformat
type
OpCode = enum opNOP, opLDA, opSTA, opADD, opSUB, opBRZ, opJMP, opSTP
Instruction = byte
Address = 0..31
# Definition of a computer zero.
Computer = object
mem: array[Address, byte] # Memory.
pc: Address # Program counter.
acc: byte # Accumulator.
Program = tuple
name: string # Name of the program.
code: seq[Instruction] # List of instructions.
proc split(inst: Instruction): tuple[code: OpCode, address: Address] =
## Split an instruction into an opCode and a value.
(OpCode((inst and 0b11100000u8) shr 5), Address(inst and 0b00011111u8))
proc toInst(opCode: OpCode; address: int): Instruction =
## Build an instruction with given opcode and value.
if address notin Address.low..Address.high:
raise newException(ValueError, "address out of range.")
byte(opCode) shl 5 or byte(address)
# Functions to build instructions.
func NOP(n: int): Instruction = toInst(opNOP, n)
func LDA(n: int): Instruction = toInst(opLDA, n)
func STA(n: int): Instruction = toInst(opSTA, n)
func ADD(n: int): Instruction = toInst(opADD, n)
func SUB(n: int): Instruction = toInst(opSUB, n)
func BRZ(n: int): Instruction = toInst(opBRZ, n)
func JMP(n: int): Instruction = toInst(opJMP, n)
func STP(n: int): Instruction = toInst(opSTP, n)
proc reset(comp: var Computer) =
## Reset the computer state.
comp.mem.reset()
comp.pc = 0
comp.acc = 0
proc load(comp: var Computer; program: Program) =
## Load a program into a computer.
comp.reset()
var idx = 0
for inst in program.code:
comp.mem[idx] = inst
inc idx
proc run(comp: var Computer) =
## Run a computer.
## The execution starts or resumes at the instruction
## at current PC, with current value in the accumulator.
while true:
let (opCode, address) = comp.mem[comp.pc].split()
inc comp.pc
case opCode
of opNOP:
discard
of opLDA:
comp.acc = comp.mem[address]
of opSTA:
comp.mem[address] = comp.acc
of opADD:
comp.acc += comp.mem[address]
of opSUB:
comp.acc -= comp.mem[address]
of opBRZ:
if comp.acc == 0:
comp.pc = address
of opJMP:
comp.pc = address
of opSTP:
break
proc enterValue(comp: var Computer; value: byte) =
## Enter a value into memory at address given by the PC,
## then increment the PC.
comp.mem[comp.pc] = value
inc comp.pc
# Programs are built using the functions NOP, LDA, STA, etc.
# To be able to use Uniform Function Call Syntax (i.e. call without parentheses),
# all these functions have an argument, even NOP and STP.
const
Prog1 = (name: "2+2",
code: @[LDA 3, ADD 4, STP 0, 2, 2])
Prog2 = (name: "7*8",
code: @[LDA 12, ADD 10, STA 12, LDA 11, SUB 13, STA 11, BRZ 8, JMP 0,
LDA 12, STP 0, 8, 7, 0, 1])
Prog3 = (name: "Fibonacci",
code: @[LDA 14, STA 15, ADD 13, STA 14, LDA 15, STA 13, LDA 16, SUB 17,
BRZ 11, STA 16, JMP 0, LDA 14, STP 0, 1, 1, 0,
8, 1])
Prog4 = (name: "List",
code: @[LDA 13, ADD 15, STA 5, ADD 16, STA 7, NOP 0, STA 14, NOP 0,
BRZ 11, STA 15, JMP 0, LDA 14, STP 0, LDA 0, 0, 28,
1, 0, 0, 0, 6, 0, 2, 26,
5, 20, 3, 30, 1, 22, 4, 24])
Prog5 = (name: "Prisoner",
code: @[NOP 0, NOP 0, STP 0, NOP 0, LDA 3, SUB 29, BRZ 18, LDA 3,
STA 29, BRZ 14, LDA 1, ADD 31, STA 1, JMP 2, LDA 0, ADD 31,
STA 0, JMP 2, LDA 3, STA 29, LDA 1, ADD 30, ADD 3, STA 1,
LDA 0, ADD 30, ADD 3, STA 0, JMP 2, 0, 1, 3])
var computer: Computer
for prog in [Prog1, Prog2, Prog3, Prog4]:
computer.load prog
computer.run()
echo &"Result for {prog.name}: {computer.acc}"
# "Prog5" is different as it stops and waits for an input from the user.
# Input is stored at address 3 (current PC value) and scores are stored at addresses 0 and 1.
type Action = enum cooperate, defect
echo &"\nResult for {Prog5.name}:"
computer.load Prog5
computer.run()
for round in 1..5:
let action = Action(round and 1)
computer.enterValue(byte(action))
computer.run()
echo &"Round {round} Action: {action:9} Player: {computer.mem[0]} Computer: {computer.mem[1]}"
</syntaxhighlight>
{{out}}
<pre>Result for 2+2: 4
Result for 7*8: 56
Result for Fibonacci: 55
Result for List: 6
Result for Prisoner:
Round 1 Action: defect Player: 0 Computer: 3
Round 2 Action: cooperate Player: 3 Computer: 3
Round 3 Action: defect Player: 3 Computer: 6
Round 4 Action: cooperate Player: 6 Computer: 6
Round 5 Action: defect Player: 6 Computer: 9
</pre>
=={{header|Phix}}==
Assumes there is no need for an "assembler", just a runtime interpreter. Output matches Algol 68.
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">constant</span> <span style="color: #000000;">NOP</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0b000_00000</span><span style="color: #0000FF;">,</span> <span style="color: #000080;font-style:italic;">-- no operation</span>
Line 1,702 ⟶ 2,914:
<span style="color: #000080;font-style:italic;">-- overflow on addition</span>
<span style="color: #000000;">execute</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"1+255"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">LDA</span><span style="color: #0000FF;">+</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">ADD</span><span style="color: #0000FF;">+</span><span style="color: #000000;">4</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">STP</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">255</span><span style="color: #0000FF;">})</span>
<!--</
You could of course define the programs using (say) 0b001_00011 instead of LDA+3.
{{out}}
Line 1,714 ⟶ 2,926:
0-1: 255
1+255: 0
</pre>
=={{header|PL/M}}==
{{works with|8080 PL/M Compiler}} ... under CP/M (or an emulator)
Uses PL/M's macro facility to provide "inline assembler" for the computerZero instructions.
The assembler code is enclosed by CODE-EDOC brackets (CODE-EDOC bracketing for inline assembler was used in some Algol 68 compilers - EDOC being CODE backwards).
<br>An additional pseudo-instruction: VAL is used to indicate literal values.
<br>As all PL/M arithmetic is unsigned and the BYTE type is 8-bits, computations are naturally MOD 255 and so there is no need to force the results to be in the range 0-255.
<syntaxhighlight lang="plm">
100H: /* EXECUTE SOME COMPUTERZERO PROGRAMS */
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* TASK */
DECLARE FALSE LITERALLY '0'
, TRUE LITERALLY '0FFH'
;
DECLARE NOP$C LITERALLY '0' /* INSTRUCTION CODES */
, LDA$C LITERALLY '32'
, STA$C LITERALLY '(2*32)'
, ADD$C LITERALLY '(3*32)'
, SUB$C LITERALLY '(4*32)'
, BRZ$C LITERALLY '(5*32)'
, JMP$C LITERALLY '(6*32)'
, STP$C LITERALLY '(7*32)'
;
DECLARE CODE LITERALLY 'PC=255' /* ASSEMBLER MNEMONICS */
, LOAD LITERALLY 'PC=PC+1;PGM(PC)='
, NOP LITERALLY 'LOAD NOP$C'
, LDA LITERALLY 'LOAD LDA$C+'
, STA LITERALLY 'LOAD STA$C+'
, ADD LITERALLY 'LOAD ADD$C+'
, SUB LITERALLY 'LOAD SUB$C+'
, BRZ LITERALLY 'LOAD BRZ$C+'
, JMP LITERALLY 'LOAD JMP$C+'
, STP LITERALLY 'LOAD STP$C '
, VAL LITERALLY 'LOAD '
, EDOC LITERALLY 'CALL EXECUTE'
;
DECLARE PGM (32)BYTE; /* THE PROGRAM */
DECLARE PC BYTE; /* PROGRAM COUNTER */
EXECUTE: PROCEDURE; /* EXECUTES THE PROGRAM */
DECLARE ( RUNNING, A, OP, OPERAND ) BYTE;
PC, A = 0;
RUNNING = TRUE;
DO WHILE RUNNING;
OP = PGM( PC ) AND 0E0H;
OPERAND = PGM( PC ) AND 1FH;
PC = ( PC + 1 ) MOD 32;
IF OP = NOP$C THEN DO; /* NOTHING */ END;
ELSE IF OP = LDA$C THEN A = PGM( OPERAND );
ELSE IF OP = STA$C THEN PGM( OPERAND ) = A;
ELSE IF OP = ADD$C THEN A = A + PGM( OPERAND );
ELSE IF OP = SUB$C THEN A = A - PGM( OPERAND );
ELSE IF OP = BRZ$C THEN DO; IF A = 0 THEN PC = OPERAND MOD 32; END;
ELSE IF OP = JMP$C THEN PC = OPERAND MOD 32;
ELSE DO; /* MUST BE STP$C */
RUNNING = FALSE;
CALL PR$NUMBER( A );
CALL PR$NL;
END;
END;
END EXECUTE ;
/* TEST PROGRAMS (FROM THE COMPUTER ZERO WEBSITE) */
CALL PR$STRING( .' 2+2: $' );
CODE; LDA 3; ADD 4; STP; VAL 2; VAL 2; EDOC;
CALL PR$STRING( .' 7*8: $' );
CODE; LDA 12; ADD 10; STA 12; LDA 11; SUB 13; STA 11; BRZ 8; JMP 0;
LDA 12; STP; VAL 8; VAL 7; VAL 0; VAL 1;
EDOC;
CALL PR$STRING( .' FIBONACCI: $' );
CODE; LDA 14; STA 15; ADD 13; STA 14; LDA 15; STA 13; LDA 16; SUB 17;
BRZ 11; STA 16; JMP 0; LDA 14; STP;
VAL 1; VAL 1; VAL 0; VAL 8; VAL 1;
EDOC;
CALL PR$STRING( .'LINKEDLIST: $' );
CODE; LDA 13; ADD 15; STA 5; ADD 16; STA 7; NOP; STA 14; NOP;
BRZ 11; STA 15; JMP 0; LDA 14; STP; LDA 0; VAL 0; VAL 28;
VAL 1; VAL 0; VAL 0; VAL 0; VAL 6; VAL 0; VAL 2; VAL 26;
VAL 5; VAL 20; VAL 3; VAL 30; VAL 1; VAL 22; VAL 4; VAL 24;
EDOC;
CALL PR$STRING( .' PRISONER: $' );
CODE; NOP; NOP; STP; VAL 0; LDA 3; SUB 29; BRZ 18; LDA 3;
STA 29; BRZ 14; LDA 1; ADD 31; STA 1; JMP 2; LDA 0; ADD 31;
STA 0; JMP 2; LDA 3; STA 29; LDA 1; ADD 30; ADD 3; STA 1;
LDA 0; ADD 30; ADD 3; STA 0; JMP 2; VAL 0; VAL 1; VAL 3;
EDOC;
/* SUBTRACTIONS YIELDING NEGATIVE RESULTS */
CALL PR$STRING( .' 0-255: $' );
CODE; LDA 3; SUB 4; STP; VAL 0; VAL 255; EDOC;
CALL PR$STRING( .' 0-1: $' );
CODE; LDA 3; SUB 4; STP; VAL 0; VAL 1; EDOC;
/* OVERFLOW ON ADDITION */
CALL PR$STRING( .' 1+255: $' );
CODE; LDA 3; ADD 4; STP; VAL 1; VAL 255; EDOC;
EOF
</syntaxhighlight>
{{out}}
<pre>
2+2: 4
7*8: 56
FIBONACCI: 55
LINKEDLIST: 6
PRISONER: 0
0-255: 1
0-1: 255
1+255: 0
</pre>
=={{header|Python}}==
<
import re
Line 1,987 ⟶ 3,329:
if __name__ == "__main__":
main()
</syntaxhighlight>
{{out}}
Line 2,000 ⟶ 3,342:
0
</pre>
=={{header|Raku}}==
Didn't bother implementing an assembler or disassembler.
<syntaxhighlight lang="raku" line>my (@mem, $ip, $acc); # memory, instruction pointer, accumulator
sub load (@program) {
die "memory exceeded" if +@program > 32;
$ip = 0;
$acc = 0;
@mem = @program »%» 256;
}
sub execute (*@program) {
load @program;
loop {
my $reg = @mem[$ip];
quietly my ($inst, $data) = $reg +> 5 +& 7, $reg +& 31;
given $inst {
when 0 { #`{ NOP } }
when 1 { #`{ LDA } $acc = @mem[$data] }
when 2 { #`{ STA } @mem[$data] = $acc }
when 3 { #`{ ADD } $acc += @mem[$data] }
when 4 { #`{ SUB } $acc -= @mem[$data] }
when 5 { #`{ BRZ } $ip = $data - 1 unless $acc }
when 6 { #`{ JMP } $ip = $data - 1 }
when 7 { #`{ STP } last }
default { die 'Invalid instruction' }
}
++$ip;
last if $ip == 32;
}
$acc
}
say "2 + 2 = " ~ execute 35,100,224,2,2;
say "7 x 8 = " ~ execute 44,106,76,43,141,75,168,192,44,224,8,7,0,1;
say "Fibonacci: " ~
execute 46,79,109,78,47,77,48,145,171,80,192,46,224,1,1,0,8,1;
say "Linked list: " ~
execute 45,111,69,112,71,0,78,0,171,79,192,46,
224,32,0,28,1,0,0,0,6,0,2,26,5,20,3,30,1,22,4,24;
say "Prisoners dilemma: " ~
execute 0,0,224,0,35,157,178,35,93,174,33,127,65,194,
32,127,64,194,35,93,33,126,99,65,32,126,99,64,194,0,1,3;</syntaxhighlight>
{{out}}
<pre>2 + 2 = 4
7 x 8 = 56
Fibonacci: 55
Linked list: 6
Prisoners dilemma: 0</pre>
=={{header|RPL}}==
{{works with|Halcyon Calc|4.2.8}}
{| class="wikitable"
! RPL code
! Comment
|-
|
{
≪ SWAP DROP ''''RAM'''' SWAP 1 + GET ≫
≪ ''''RAM'''' SWAP 1 + 3 PICK PUT ≫
≪ ''''RAM'''' SWAP 1 + GET + 256 MOD ≫
≪ ''''RAM'''' SWAP 1 + GET - 256 MOD ≫
≪ '''IF''' OVER NOT '''THEN''' ROT DROP SWAP '''ELSE''' DROP '''END''' ≫
≪ ROT DROP SWAP ≫
≪ DROP 1 SF ≫
} ''''µCODE'''' STO
≪
{ 32 } RDM ''''RAM'''' STO
0 0 1 CF
'''WHILE''' 1 FC? '''REPEAT'''
SWAP 1 + DUP 32 MOD ROT ROT ''''RAM'''' SWAP GET
32 DUP2 MOD ROT ROT / FLOOR
'''IF THEN 'µCODE'''' LAST GET EVAL '''ELSE''' DROP
'''END END'''
≫ ‘'''RUN'''’ STO
|
'''µCODE''' ''( counter accumulator address -- counter accumulator )''
LDA x
STA x
ADD x
SUB x
BRZ x
JMP x
STP x
'''RUN''' ''( [ program ] -- counter accumulator )''
load program to memory
reset accumulator, counter and flag
loop until STP
increment pointer and read byte
instruction, addresss = divmod(byte,32)
if instruction <> NOP then execute it
|}
{{in}}
<pre>
[ 35 100 224 2 2 ] RUN
[ 44 106 76 43 141 75 168 192 44 224 8 7 0 1 ] RUN
[ 46 79 109 78 47 77 48 145 171 80 192 46 224 1 1 0 8 1 ] RUN
[ 45 111 69 112 71 0 78 0 171 79 192 46 224 32 0 28 1 0 0 0 6 0 2 26 5 20 3 30 1 22 4 24 ] RUN
[ 0 0 224 0 0 35 157 178 35 93 174 33 127 65 194 32 127 64 192 35 93 33 126 99 ] RUN
</pre>
{{out}}
<pre>
5: 4
4: 56
3: 55
2: 6
1: 0
</pre>
On a basic HP-28S, it takes 9.5 seconds takes to multiply 7 by 8 and 15 seconds to get the 11th Fibonacci number.
=={{header|Wren}}==
Line 2,005 ⟶ 3,462:
Although I've entered the fifth program, it just returns 0 when reaching the STP instruction. I'm unclear how we can make it interactive without having another instruction though possibly, if STP had address bits other than 0, this could signal that user input was required.
<
var LDA = 1
var STA = 2
Line 2,181 ⟶ 3,638:
"""
]
for (prog in progs) interp.call(prog)</
{{out}}
<pre>
4
56
55
6
0
</pre>
=={{header|XPL0}}==
Some simplifying assumptions are made. Literal addresses are used instead
of labels. The STP instruction merely stops (and displays the
accumulator) rather than allowing the program to continue after a key is
struck. This prevents the Prisoner's Dilemma from running as intended.
<syntaxhighlight lang "XPL0">def NOP=0, LDA=1, STA=2, ADD=3, SUB=4, BRZ=5, JMP=6, STP=7;
char Mem(32);
string 0;
proc Load(Prog); \Load program into memory
char Prog;
int PC, I, Ch, Hash;
func GetNum; \Return value of number
int Num;
[while Ch <= $20 do
[Ch:= Prog(I); I:= I+1];
Num:= 0;
while Ch>=^0 and Ch<=^9 do
[Num:= Num*10 + Ch - ^0;
Ch:= Prog(I); I:= I+1;
];
return Num;
];
[PC:= 0; I:= 0;
repeat repeat Ch:= Prog(I); I:= I+1; until Ch > $20;
if Ch>=^0 and Ch<=^9 then
Mem(PC):= GetNum
else [Hash:= 0;
while Ch >= ^A do
[Hash:= Hash + Ch;
Ch:= Prog(I); I:= I+1;
];
case Hash of
^N+^O+^P: Mem(PC):= NOP<<5;
^L+^D+^A: Mem(PC):= LDA<<5 + GetNum;
^S+^T+^A: Mem(PC):= STA<<5 + GetNum;
^A+^D+^D: Mem(PC):= ADD<<5 + GetNum;
^S+^U+^B: Mem(PC):= SUB<<5 + GetNum;
^B+^R+^Z: Mem(PC):= BRZ<<5 + GetNum;
^J+^M+^P: Mem(PC):= JMP<<5 + GetNum;
^S+^T+^P: Mem(PC):= STP<<5
other [Text(0, "Bad opcode at "); IntOut(0, PC); exit];
];
PC:= PC+1;
until Ch = 0;
];
proc Run(Prog); \Load and run a program
int Prog, PC, Acc, Instr, Addr;
[Load(Prog);
PC:= 0; Acc:= 0;
loop [Instr:= Mem(PC) >> 5;
Addr:= Mem(PC) & $1F;
PC:= PC+1;
case Instr of
LDA: Acc:= Mem(Addr);
STA: Mem(Addr):= Acc;
ADD: Acc:= Acc + Mem(Addr) & $FF;
SUB: Acc:= Acc - Mem(Addr) & $FF;
BRZ: if Acc = 0 then PC:= Addr;
JMP: PC:= Addr;
STP: quit
other []; \NOP
if PC > 31 then quit;
];
IntOut(0, Acc); CrLf(0);
];
int Progs, N;
[Progs:= [
"LDA 3 ADD 4 STP 2 2",
"LDA 12 ADD 10 STA 12 LDA 11 SUB 13 STA 11 BRZ 8 JMP 0 LDA 12 STP 8 7 0 1",
"LDA 14 STA 15 ADD 13 STA 14 LDA 15 STA 13 LDA 16 SUB 17 BRZ 11 STA 16 JMP 0
LDA 14 STP 1 1 0 8 1",
"LDA 13 ADD 15 STA 5 ADD 16 STA 7 NOP STA 14 NOP BRZ 11 STA 15 JMP 0 LDA 14
STP LDA 0 0 28 1 0 0 0 6 0 2 26 5 20 3 30 1 22 4 24",
"0 0 STP NOP LDA 3 SUB 29 BRZ 18 LDA 3 STA 29 BRZ 14 LDA 1 ADD 31 STA 1
JMP 2 LDA 0 ADD 31 STA 0 JMP 2 LDA 3 STA 29 LDA 0 ADD 30 ADD 3 STA 0 LDA 1
ADD 30 ADD 3 STA 1 JMP 2 0 1 3"];
for N:= 0 to 4 do Run(Progs(N));
]</syntaxhighlight>
{{out}}
<pre>
Line 2,194 ⟶ 3,743:
=={{header|Z80 Assembly}}==
Output is in hexadecimal but is otherwise correct.
<
PrintChar equ &BB5A
Line 2,388 ⟶ 3,937:
db %00000000,%00000000
db %00000000,%00000000
db %00000000,%00000000</
{{out}}
<pre>04
| |||