Implement the Universal Machine emulator

Question

The goal is to write a full program that emulates the Universal Machine from ICFP 2006 with the shortest code. The Universal Machine has a very simple instruction set explained here. The emulator has to read a filename from command-line argument, and run the file as the program, so your language has to support command-line arguments and stdin/out in some way. The emulator has to complete the sandmark within a reasonable time (not decades). Here's a short explanation of the instruction set:

The machine has eight registers each holding an 32-bit unsigned integer.
The machine holds an indexed set of arrays of 32-bit unsigned integer cells.
Shortly speaking, the allocation instruction returns an opaque 32-bit uint which is the handle to the created array, which has a static size, and holds 32-bit uint elements.
The 0'th array denotes the program. It is loaded from a big-endian file on startup.
There is also an Instruction Pointer which points to a cell in the 0 array.
On each step, an instruction is read from the cell the Pointer points to, and the Pointer is inceremented before anything is done.
The 4 most significant bits represent the opcode.
If the opcode is 13, then the next 3 bits represent the register, and the other 25 represent the number that is written into the said register.
Otherwise the 9 least significant bits represent three registers, say, A, B, and C, where C is represented by the 3 least significant bits.
Then depending on the opcode, the following happens:
0. A = B unless C == 0
1. A = B[C]
2. A[B] = C
3. A = B + C
4. A = B * C
5. A = B / C
6. A = ~(B & C)
7. The emulator exits
8. B = allocate(C)
9. deallocate(C)
10. output a character from C to stdout
11. input a character from stdin into C
12. copy the array B into the 0 array and set the Pointer to C

I have written a needlessly complex but totally fast implementation (ab)using x86_64 jitted assembly (the fun starts in emit()), which would for sure help you if you misunderstand some of the Machine's aspects.

Answer 1

PHP: 443 416  384 bytes

<?php @eval(ereg_replace('[U-Z]','$\0',strtr('for(Y=[unpack("N*",join(file($argv[1])))];;A|=0){{W=Y[V=0][++U]
C&&A=B
A=Y[B][C+1]
Y[A][B+1]=C
A=B+C
A=B*C
A=bcdiv(PB),PC))*1
A=~B|~C
die
B=++Z
unset(Y[C])
echo chr(C)
C=fgetc(STDIN);C=ord(C)-(C=="")
Y[0]=Y[B|0];U=C
X[W>>25&7]=W&33554431;}}',['
'=>';}if((W>>28&15)==V++){',A=>'X[W>>6&7]',B=>'X[W>>3&7]',C=>'X[W&7]',P=>'sprintf("%u",'])));

*Revamped again*. It's as small as I can possibly get it now. I kept some variables down the far end of the alphabet so that the regex that inserts the $ signs doesn't mangle the STDIN constant, so here's a little glossary:

• U: instruction pointer
• V: index of opcode that's currently being tested
• W: current instruction word
• X: the 8 general purpose registers
• Y: main memory (each block is 1-based, since that's how unpack() returns arrays)
• Z: id of next free memory block (will eventually overflow, but the sandmark only uses ~92 million)
• A,B,C are the registers of the current instruction as in the spec

Unsigned division is a subtle nuisance (the *1 is needed to ensure that large numbers cast back to the correct int) but the rest of the arithmetic is easy to keep 32-bit by ORing the arithmetic register with 0 (A|=0) after each instruction.

---

I found this project really interesting but striving to minimize the character count made it slow and inelegant, so I also made a simple (not golfed) Java version, that can complete the sandmark in a few minutes instead of taking all day:

import java.io.*;
import java.util.HashMap;

public class UniversalMachine {
    public static void main(String[] args) throws IOException {
        if (args.length == 0) {
            System.err.println("Program not specified.");
            System.exit(1);
        }

        int[] program;
        try (RandomAccessFile raf = new RandomAccessFile(args[0], "r")) {
            program = new int[(int)(raf.length() / 4)];
            for (int i = 0; i < program.length; i++) {
                program[i] = raf.readInt();
            }
        }

        HashMap<Integer,int[]> memory = new HashMap<>();
        memory.put(0, program);
        int nextMemKey = 1;

        int[] R = new int[8]; // Registers
        int IP = 0; // Execution Finger (Instruction Pointer)

        loop: for (;;) {
            int ins = program[IP++];
            int op = ins >>> 28;
            if (op == 13) { // Orthography
                int A = (ins >> 25) & 7;
                int num = ins & 0x01FF_FFFF;
                R[A] = num;
            } else {
                final int A = (ins >> 6) & 7;
                final int B = (ins >> 3) & 7;
                final int C = (ins >> 0) & 7;
                switch (op) {
                case 0: // Conditional Move
                    if (R[C] != 0) R[A] = R[B];
                    break;
                case 1: // Array Index
                    R[A] = memory.get(R[B])[R[C]];
                    break;
                case 2: // Array Amendment
                    memory.get(R[A])[R[B]] = R[C];
                    break;
                case 3: // Addition
                    R[A] = R[B] + R[C];
                    break;
                case 4: // Multiplication
                    R[A] = R[B] * R[C];
                    break;
                case 5: // Division
                    R[A] = (int)((R[B] & 0xFFFF_FFFFL) / (R[C] & 0xFFFF_FFFFL));
                    break;
                case 6: // Not-And
                    R[A] = ~(R[B] & R[C]);
                    break;
                case 7: // Halt
                    break loop;
                case 8: // Allocation
                    // note: must use C before setting B, as they may be the same reg
                    memory.put(nextMemKey, new int[R[C]]);
                    R[B] = nextMemKey++;
                    break;
                case 9: // Abandonment
                    memory.remove(R[C]);
                    break;
                case 10: // Output
                    System.out.print((char)R[C]);
                    break;
                case 11: // Input
                    R[C] = System.in.read();
                    break;
                case 12: // Load Program
                    IP = R[C];
                    if (R[B] != 0) {
                        memory.put(0, program = memory.get(R[B]).clone());
                    }
                    break;
                }
            }
        }
    }
}

Answer 2

Perl, 407

It looks like the question might seem too complex, actually it's very simple.
I'm still very new to perl though, anyway here it is

open$f,shift;binmode$f;push@{$m[0]},unpack'N',$b while read$f,$b,4;$z=2**32;while(){$o=$m[0][$p++];$a=\$r[$o>>6&7];$b=\$r[$o>>3&7];$c=\$r[$o&7];eval qw,$$a=($$b)if$$c $$a=$m[$$b][$$c] $m[$$a][$$b]=$$c $$a=($$b+$$c)%$z $$a=$$b*$$c%$z $$a=$==$$b/$$c $$a=$$b&$$c^($z-1) exit $$b=scalar@m;$m[$$b]=[] undef$m[$$c] print(chr$$c) $$c=ord(getc) $m[0]=[@{$m[$$b]}]if$$b;$p=$$c $r[$o>>25&7]=$o&33554431,[$o>>28].";";}

It runs really slow, probably 800x slower than the JITed x86_64 one.
Also, a friend of mine made a reference C implementation

Answer 3

C, 924 838 825 696 646 623

I store a "pointer" (byte-offset) in the register designated b in the instruction, and use whatever register designates an array in the pseudocode in the same manner (or reverse, rather, to reconstitute a pointer) to access that array later. Still need to try the test program...

Edit: added comments.

Edit: fixed instruction 12. change the pointer, not the instruction in memory. Count is with all comments, indents, and newlines removed.

Edit: It appears to be running now, assuming I'm interpreting the results correctly. :) The final realization was that the 0 array is indeed referenced by the handle 0, which may be found in an uninitialized register. A very twisted little machine! :)

Edit: rewrote debugging apparatus to use write instead of printf.... The idea here is to remove bugs. :) Edit: putchar() and getchar() are also no-nos with sbrk. It now works, and appears quite fast.

#define O(_)*a=*b _*c;B
#define B break;case
#define U unsigned
U*m,r[8],*p,*z,f,x,*a,*b,*c;main(int n,char**v){U char
u[4];z=m=p=sbrk(4);f=n>1?open(v[1],0):0;\
while(read(f,u,4)){*m++=(((((*u<<8)|u[1])<<8)|u[2])<<8)|u[3];sbrk(4);}sbrk(4);\
for(;x=*p++,1;){c=r+(x&7);b=r+((x>>3)&7);a=r+((x>>6)&7);switch(x>>28){case
0:*c?*a=*b:0;B
1:*a=(*b?m+*b:z)[*c];B
2:(*a?m+*a:z)[*b]=*c;B
3:O(+)4:O(*)5:O(/)6:*a=~(*b&*c);B
7:return 0;case
8:*b=1+(U*)sbrk(4*(1+*c))-m;(m+*b)[-1]=*c;B
9:B
10:*u=*c;write(1,u,1);B 
11:read(0,u,1);*c=*u;B
12:*b?memcpy(z=sbrk(4*(m+*b)[-1]),m+*b,4*(m+*b)[-1]):0;p=&z[*c];B
13:a=r+((x>>25)&7);*a=x&0x1ffffff;}}}

For little-endian only, there's a 611 character version.

#define O(_)*a=*b _*c;B
#define B break;case
#define U unsigned
U*m,r[8],*p,*z,f,x,*a,*b,*c;main(int n,char**v){U char
u[4];z=m=p=sbrk(4);f=n>1?open(v[1],0):0;while(read(f,u,4)){*m++=(((((*u<<8)|u[1])<<8)|u[2])<<8)|u[3];sbrk(4);}sbrk(4);for(;x=*p++,1;){c=r+(x&7);b=r+((x>>3)&7);a=r+((x>>6)&7);switch(x>>28){case
0:*c?*a=*b:0;B
1:*a=(*b?m+*b:z)[*c];B
2:(*a?m+*a:z)[*b]=*c;B
3:O(+)4:O(*)5:O(/)6:*a=~(*b&*c);B
7:return 0;case
8:*b=1+(U*)sbrk(4*(1+*c))-m;(m+*b)[-1]=*c;B
9:B
//10:*u=*c;write(1,u,1);B //generic
10:write(1,c,1);B //little-endian
//11:read(0,u,1);*c=*u;B //generic
11:read(0,c,1);B //little-endian
12:*b?memcpy(z=sbrk(4*(m+*b)[-1]),m+*b,4*(m+*b)[-1]):0;p=&z[*c];B
13:a=r+((x>>25)&7);*a=x&0x1ffffff;}}}

Indented and commented, with (extended) commented debugging apparatus.

//#define DEBUG 1
#include <fcntl.h> // open
#include <signal.h> // signal
#include <stdio.h> // putchar getchar
#include <string.h> // memcpy
#include <sys/types.h> // open
#include <sys/stat.h> // open
#include <unistd.h> // sbrk read
unsigned long r[8],*m,*p,*z,f,x,o,*a,*b,*c; // registers memory pointer zero file working opcode A B C
char alpha[] = "0123456789ABCDEF";
//void S(int x){signal(SIGSEGV,S);sbrk(9);} // autogrow memory while reading program
void writeword(int fd, unsigned long word){
    char buf[8];
    unsigned long m=0xF0000000;
    int off;
    for (off = 28; off >= 0; m>>=4, off-=4) {
        buf[7-(off/4)]=alpha[(word&m)>>off];
    }
    write(fd, buf, 8);
    write(fd, " ", 1);
}
int main(int n,char**v){
#ifdef DEBUG
    int fdlog;
#endif
    unsigned char u[4]; // 4-byte buffer for reading big-endian 32bit words portably
    int cnt;

#ifdef DEBUG
    fdlog = open("sandlog",O_WRONLY|O_CREAT|O_TRUNC, 0777);
#endif
    z=m=p=sbrk(4); // initialize memory and pointer
    //signal(SIGSEGV,S); // invoke autogrowing memory -- no longer needed
    f=n>1?open(v[1],O_RDONLY):0; // open program
    while(read(f,u,4)){ // read 4 bytes
        *m++=(((((*u<<8)|u[1])<<8)|u[2])<<8)|u[3]; // pack 4 bytes into 32bit unsigned in mem
        sbrk(4); // don't snip the end of the program
    }
    sbrk(4);
    for(cnt=0;x=*p++,1;cnt++){ // working = *ptr; ptr+=1
        c=r+(x&7); // interpret C register field
        b=r+((x>>3)&7); // interpret B register field
        a=r+((x>>6)&7); // interpret A register field
#ifdef DEBUG
        {int i;write(fdlog,"{",1);for(i=0;i<8;i++)writeword(fdlog, r[i]);
            write(fdlog,"} ",2);
        }
        write(fdlog, alpha+(x), 1);
        write(fdlog, alpha+(x>>28), 1);
#endif
        switch(o=x>>28){ // interpret opcode
            case 0:
#ifdef DEBUG
                write(fdlog, "if(rX)rX=rX\n", 12);
#endif
                *c?*a=*b:0;
                break; // Conditional Move A=B unless C==0
            case 1:
#ifdef DEBUG
                write(fdlog, "rX=rX[rX]\n", 10);
#endif
                *a=(*b?m+*b:z)[*c];
                break; // Array Index A=B[C]
            case 2:
#ifdef DEBUG
                write(fdlog, "rX[rX]=rX\n", 10);
#endif
                (*a?m+*a:z)[*b]=*c;
                break; // Array Amendment A[B] = C
            case 3:
#ifdef DEBUG
                write(fdlog, "rX=rX+rX\n", 9);
#endif
                *a=*b+*c;
                break; // Addition A = B + C
            case 4:
#ifdef DEBUG
                write(fdlog, "rX=rX*rX\n", 9);
#endif
                *a=*b**c;
                break; // Multiplication A = B * C
            case 5:
#ifdef DEBUG
                write(fdlog, "rX=rX/rX\n", 9);
#endif
                *a=*b/ *c;
                break; // Division A = B / C
            case 6:
#ifdef DEBUG
                write(fdlog, "rX=~(rX&rX)\n", 12);
#endif
                *a=~(*b&*c);
                break; // Not-And A = ~(B & C)
            case 7:
#ifdef DEBUG
                write(fdlog, "halt\n", 5);
#endif
                return 0; // Halt 
            case 8:
#ifdef DEBUG
                write(fdlog, "rX=alloc(rX)\n", 13);
#endif
                *b=1+(unsigned long*)sbrk(4*(1+*c))-m;
                   (m+*b)[-1]=*c;

                   break; // Allocation B = allocate(C)
            case 9:
#ifdef DEBUG
                   write(fdlog, "free(rX)\n", 9);
#endif
                   break; // Abandonment deallocate(C)
            case 10:
#ifdef DEBUG
                   write(fdlog, "output(rX)\n", 11);
#endif
                   //putchar(*c);
                   //*u=u[1]=u[2]=' ';
                   u[3]=(char)*c;
                   write(fileno(stdout), u+3, 1);
                   break; // Output char from C to stdout
            case 11:
#ifdef DEBUG
                   write(fdlog, "rX=input()\n", 11);
#endif
                   //x=getchar();*c=x;
                   read(fileno(stdin), u+3, 1);
                   *c=u[3];
                   break; // Input char from stdin into C
            case 12:
#ifdef DEBUG
                   write(fdlog, "load(rX)[rX]\n", 13);
#endif
                    *b?memcpy(z=sbrk(4*(m+*b)[-1]),m+*b,4*(m+*b)[-1]):0;
                    p=&z[*c];
                    break; // Load Program copy the array B into the 0 array, Ptr=C
            case 13:
#ifdef DEBUG
                    write(fdlog, "rX=X\n", 5);
#endif
                    a=r+((x>>25)&7);*a=x&0x1ffffff; // Orthography REG=immediate-25bit
        }
    }
}