C, 924 838 825 696 646 623
Saya menyimpan "pointer" (byte-offset) dalam register yang ditunjuk bdalam instruksi, dan menggunakan register apa pun yang menunjuk array dalam pseudocode dengan cara yang sama (atau membalikkan, lebih tepatnya, untuk menyusun kembali pointer) untuk mengakses array itu nanti. Masih perlu mencoba program tes ...
Edit: komentar ditambahkan.
Sunting: instruksi tetap 12. ubah pointer, bukan instruksi di memori. Hitung dengan semua komentar, indentasi, dan baris baru dihapus.
Sunting: Tampaknya sedang berjalan sekarang, dengan asumsi saya menafsirkan hasil dengan benar. :) Realisasi terakhir adalah bahwa 0 array memang direferensikan oleh handle 0, yang dapat ditemukan di register yang tidak diinisialisasi. Mesin kecil yang sangat bengkok! :)
Sunting: menulis ulang alat debugging untuk digunakan writealih-alih printf.... Idenya di sini adalah untuk menghapus bug. :) Edit: putchar() dan getchar()juga no-no dengan sbrk. Sekarang berfungsi, dan muncul cukup cepat.
#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;}}}
Untuk little-endian saja, ada versi 611 karakter.
#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;}}}
Diindentasi dan dikomentari, dengan (diperpanjang) berkomentar alat debugging.
//#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
}
}
}