Hi All,
I am attempting to reverse-obsfuscate an IOCCC entry, and my ultimate
goal is to convert the code into Delphi (OO Pascal).
I don't understand what the question-mark (?) operator, and the
percent (%) operator does.
How do I translate this line (obviously declaring some integers) into
pascal ???
int c = 0, *n = v + 100, j = d < u - 1 ? a : - 9000, w, z, i, g, q
= 3-f;
And how do I translate these lines to pascal ???
for(w = i = 0; i < 4; i++)
w += (m = v[h[i]]) == f ? 300 : m == q ? - 300: (t = v[ih[i]])
== f ? - 50: t == q ? 50 : 0;
And how do I translate these lines to pascal ???
for(; t < 1600; t += 100)
for(m = 0; m < 100; m++)
V[t + m] = m < 11 || m > 88 ||(m + 1) % 10 < 2 ? 3 : 0;
What does "<<" do ? e.g. how do I translate the next line???
return d >= u - 1 ? j + (c << 3): j;
Any help will be appreciated.
Regards,
Peter W. :-)))
Sandy Bay, Hobart, Tas, AU.
obsfuscated code ("lievaart2.c")
-----
#define D define
#D Y return
#D R for
#D e while
#D I printf
#D l int
#D W if
#D C y=v+111;H(x,v)*y++= *x
#D H(a,b)R(a=b+11;a<b+89;a++)
#D s(a)t=scanf("%d",&a)
#D U Z I
#D Z I("123\
45678\n");H(x,V){putchar(".XO"[*x]);W((x-V)%10==8){x+=2;I("%d\n",(x-V)/1
0-1);}}
l
V[1600],u,r[]={-1,-11,-10,-9,1,11,10,9},h[]={11,18,81,88},ih[]={22,27,72
,77},
bz,lv=60,*x,*y,m,t;S(d,v,f,_,a,b)l*v;{l
c=0,*n=v+100,j=d<u-1?a:-9000,w,z,i,g,q=
3-f;W(d>u){R(w=i=0;i<4;i++)w+=(m=v[h[i]])==f?300:m==q?-300:(t=v[ih[i]])=
=f?-50:
t==q?50:0;Y w;}H(z,0){W(E(v,z,f,100)){c++;w=
-S(d+1,n,q,0,-b,-j);W(w>j){g=bz=z;
j=w;W(w>=b||w>=8003)Y w;}}}W(!c){g=0;W(_){H(x,v)c+=
*x==f?1:*x==3-f?-1:0;Y c>0?
8000+c:c-8000;}C;j= -S(d+1,n,q,1,-b,-j);}bz=g;Y
d>=u-1?j+(c<<3):j;}main(){R(;t<
1600;t+=100)R(m=0;m<100;m++)V[t+m]=m<11||m>88||(m+1)%10<2?3:0;I("Level:"
);V[44]
=V[55]=1;V[45]=V[54]=2;s(u);e(lv>0){Z
do{I("You:");s(m);}e(!E(V,m,2,0)&&m!=99);
W(m!=99)lv--;W(lv<15&&u<10)u+=2;U("Wait\n");I("Value:%d\n",S(0 ,V,1,0,-90
00,9000
));I("move:
%d\n",(lv-=E(V,bz,1,0),bz));}}E(v,z,f,o)l*v;{l*j,q=3-f,g=0,i,w,*k=v
+z;W(*k==0)R(i=7;i>=0;i--){j=k+(w=r[i]);e(*j==q)j+=w;W(*j==f&&j-w!=k){W(
!g){g=1
;C;}e(j!=k)*((j-=w)+o)=f;}}Y g;}
-----
un-obsfuscated code:
-----
/* This source has been formatted by an unregistered SourceFormatX */
/* If you want to remove this info, please register this shareware */
/* Please visit http://www.textrush.com to get more information */
H(x,V)
{
putchar(".XO"[*x]);
if((x-V)%10==8)
{
x+=2;
printf("%d\n",(x-V)/10-1);
}
}
int V[1600], u, r[] =
{
- 1, - 11, - 10, - 9, 1, 11, 10, 9
}
, h[] =
{
11, 18, 81, 88
}
, ih[] =
{
22, 27, 72, 77
}
, bz, lv = 60, *x, *y, m, t;
S(d, v, f, _, a, b)int *v;
{
int c = 0, *n = v + 100, j = d < u - 1 ? a : - 9000, w, z, i, g, q
= 3-f;
if(d > u)
{
for(w = i = 0; i < 4; i++)
w += (m = v[h[i]]) == f ? 300 : m == q ? - 300: (t = v[ih[i]])
== f ? - 50: t == q ? 50 : 0;
return w;
}
H(z, 0)
{
if(E(v, z, f, 100))
{
c++;
w = - S(d + 1, n, q, 0, - b, - j);
if(w > j)
{
g = bz = z;
j = w;
if(w >= b || w >= 8003)return w;
}
}
}
if(!c)
{
g = 0;
if(_)
{
H(x, v)c += *x == f ? 1 : *x == 3-f ? - 1: 0;
return c > 0 ? 8000+c: c - 8000;
}
y=v+111;
H(x,v)*y++= *x;
j = - S(d + 1, n, q, 1, - b, - j);
}
bz = g;
return d >= u - 1 ? j + (c << 3): j;
}
main()
{
for(; t < 1600; t += 100)
for(m = 0; m < 100; m++)
V[t + m] = m < 11 || m > 88 ||(m + 1) % 10 < 2 ? 3 : 0;
printf("Level:");
V[44] = V[55] = 1;
V[45] = V[54] = 2;
s(u);
while(lv > 0)
{
printf("123\45678\n");
do
{
printf("You:");
s(m);
}
while(!E(V, m, 2, 0) && m != 99);
if(m != 99)lv--;
if(lv < 15 && u < 10)u += 2;
printf("123\45678\n");
printf("Wait\n");
printf("Value:%d\n", S(0, V, 1, 0, - 9000, 9000));
printf("move: %d\n", (lv -= E(V, bz, 1, 0), bz));
}
}
E(v, z, f, o)int *v;
{
int *j, q = 3-f, g = 0, i, w, *k = v + z;
if(*k == 0)
for(i = 7; i >= 0; i--)
{
j = k + (w = r[i]);
while(*j == q)j += w;
if(*j == f && j - w != k)
{
if(!g)
{
g = 1;
y=v+111;
H(x,v)*y++= *x;
}
while(j != k)*((j -= w) + o) = f;
}
}
return g;
}
-----
lievaart.hint
-----
Grand Prize:
Roemer B. Lievaart
VU Informatica
Churchilllaan 173-IV
Amsterdam, The Netherlands
We believe that you too will be amazed at just how much power Mr.
Lievaart packed into 1024 bytes!
This Plays the game of reversi (Othello)! Compile and run. It then
asks for a playing level. Enter 0-10 (easy-hard). It then asks for
your move. A move is a number within 11-88, or a 99 to pass. Illegal
moves (except for an illegal pass) are rejected. Then the computer
does its move (or a 0 to pass), until the board is full. It plays
rather well, for such a small program! Lievaart had to leave out the
board printing routine, so you'll have to take a real game board to
play it. ... Also due to space-limitations (the rules for 1987 had a
limit of 1024 byes), Lievaart took out the passing-handler, which
makes its ending-game rather poor. But further it knows all the
rules, uses alpha-beta pruning, and it plays f.i. on mobility(!).
Most important: it can play a pretty good game of Reversi!
The Author was kind enough to supply the fully functional version of the
program. The file lievaart2.c contains what the program would have
been without the size restriction. This version has the full end game
logic and displays the board after each move!
Copyright (c) 1987, Landon Curt Noll & Larry Bassel.
All Rights Reserved. Permission for personal, educational or
non-profit use is granted provided this this copyright and notice are
included in its entirety and remains unaltered. All other uses must
receive prior permission in writing from both Landon Curt Noll and
Larry Bassel.
-----
