Hello,
I’m going to participate in a debugging competition going to be conducted in our college. The problems will be in C programming language. I’ve completed the problems given in our text book and am looking for other problems online.
Also I should mention that files are off topic for this competition.
Any tips for the competition are also welcome.
Thanks.
Debugging?
#include "SDL.h" #define $ for(O=9 #define CX M+=(T%3+2*!(!T*t-6)) #define x ,A=4*!T,O=t,W=h=T<3?u(Q?p:D(A+3),D(A),D(A+1)[i]+D(A+2)*g+):K(t),U=V=K(a),o?U=h,W=V:V, #define C 8*-~L #define Z short #define y a(Z)Y[++O] #define B ),a--||( #define _ ),e--||( #define V(I,D,E)(O=a(I)h[r])&&!(A=(D)(V=(1[E+L]<<16)+*i)/O,A-(I)A)?1[E+L]=V-O*(*E=A):H(0) #define i(B,M)B(o){return M;} #define R(O,M,_)(S=L?a(I Z)O:O,N=L?a(I Z)O M(f=a(I Z)_):(O M(f=a(I n)_))) #define T(_)R(r[u(10,L=4,--)],=,_) #define u(a,r,T)16*i[a]+(I Z)(T i[r]) #define a(_)*(_*)& #define L(_)M(W,_,U) #define M(S,F,T)R(r[S],F,r[T]) #define A(_)(i[L=4]+=2,R(_,=,r[u(10,4,-2+)])) #define c(R,T)(1[u=19,L+T]=(N=a(R)h[r]*(R)*T)>>16,*i=N,G(F(N-(R)N))) #define h(_)(1&(L?a(Z)_:_)>>C-1) #define I unsigned #define n char #define e(_)v(F(40[L(_##=40[E]+),E]&N==S|_ N<_(int)S)) I n t,e,l[80186],*E,m,u,L,a,T,o,r[1<<21],X,*Y,b,Q=0,R=0;I Z*i,M,p,q=3;I*localtime(),f,S,kb=0,h,W,U,c,g,d,V,A;N,O,P=983040,j[5];SDL_Surface*k=0;i(K,P+(L?2*o:2*o+o/4&7))i(D,r[a(I)E[259+4*o]+O])i(w,i[o]+=~(-2*47[E])*~L)i(v,(z((f^=S^N)&16),G(N-S&&1&(40[E]^f>>C-1))))J(){V=61442;$;O--;)V+=40[E+O]<9|42[E])),*E&=15))i(SP,(w(7),R&&--1[i]&&o?R++,Q&&Q++,M--:0))DX(){$,O*=27840;O--;)O[(I*)k->pixels]=-!!(1<<7-O%8&r[O/2880*90+O%720/8+(88+952[l]/128*4+O/720%4<<13)]);SDL_Flip(k);}main(BX,nE)n**nE;{9[i=E=r+P]=P>>4;$;q;)j[--q]=*++nE?open(*nE,32898):0;read(2[a(I)*i=*j?lseek(*j,0,2)>>9:0,j],E+(M=256),P);$;Y=r+16*9[i]+M,Y-r;Q|R||kb&46[E]&&KB)--64[T=1[O=32[L=(X=*Y&7)&1,o=X/2&1,l]=0,t=(c=y)&7,a=c/8&7,Y]>>6,g=~-T?y:(n)y,d=BX=y,l],!T*t-6&&T-2?T-1?d=g:0:(d=y),Q&&Q--,R&&R--x(O=*Y,O=u=D(51),e=D(8),m=D(14)_ O=*Y/2&7,M+=(n)c*(L^(D(m)[E]|D(22)[E]|D(23)[E]^D(24)[E]))_ L=*Y&8,R(K(X)[r],=,c)_ L=e+=3,o=0,a=X x a=m _ T(X[i])_ A(X[i])_ a<2?M(U,+=1-2*a+,P+24),v(f=1),G(S+1-a==1<S)B L?c(I Z,i):c(I/**/n,E)B L?c(Z,i):c(n,E)B L?V(I Z,I,i):V(I n,I Z,E)B L?V(Z,int,i):V(n,Z,E))_++e,h=P,d=c,T=3,a=m,M--_++e,13[W=h,i]=(o|=!L)?(n)d:d,U=P+26,M-=~!o,u=17+(m=a)_(a=m B L(+=),F(NS)B L(^=)B L(-),F(N>S)B L(=))_!L?L=a+=8 x L(=):!o?Q=1,R(r[p=m x V],=,h):A(h[r])_ T=a=0,t=6,g=c x M(U,=,W)_(A=h(h[r]),V=m?++M,(n)g:o?31&2[E]:1)&&(a<4?V%=a/2+C,R(A,=,h[r]):0,a&1?R(h[r],>>=,V):R(h[r],<<=,V),a>3?u=19:0,a<5?0:F(S>>V-1&1)B R(h[r],+=,A>>C-V),G(h(N)^F(N&1))B A&=(1<>1+C-V),G(h(N)^F(A&1<>V)))_(V=!!--1[a=X,i]B V&=!m[E]B V&=m[E]B 0 B V=!++1[i]),M+=V*(n)c _ M+=3-o,L?0:o?9[M=0,i]=BX:T(M),M+=o*L?(n)c:c _ M(U,&,W)_ L=e+=8,W=P,U=K(X)_!R||1[i]?M(m<2?u(8,7,):P,=,m&1?P:u(Q?p:11,6,)),m&1||w(6),m&2||SP(1):0 _!R||1[i]?M(m?P:u(Q?p:11,6,),-,u(8,7,)),43[u=92,E]=!N,F(N>S),m||w(6),SP(!N==b):0 _ o=L,A(M),m&&A(9[i]),m&2?s(A(V)):o||(4[i]+=c)_ R(U[r],=,d)_ 986[l]^=9,R(*E,=,l[m?2[i]:(n)c])_ R(l[m?2[i]:(n)c],=,*E)_ R=2,b=L,Q&&Q++_ W-U?L(^=),M(U,^=,W),L(^=):0 _ T(m[i])_ A(m[i])_ Q=2,p=m,R&&R++_ L=0,O=*E,F(D(m+=3*42[E]+6*40[E])),z(D(1+m)),N=*E=D(m-1)_ N=BP(m-1)_ 1[E]=-h(*E)_ 2[i]=-h(*i)_ 9[T(9[i]),T(M+5),i]=BX,M=c _ J(),T(V)_ s(A(V))_ J(),s((V&~m)+1[E])_ J(),1[E]=V _ L=o=1 x L(=),M(P+m,=,h+2)_++M,H(3)_ M+=2,H(c&m)_++M,m[E]&&H(4)_(c&=m)?1[E]=*E/c,N=*E%=c:H(0)_*i=N=m&E[L=0]+c*1[E]_*E=-m[E]_*E=r[u(Q?p:m,3,*E+)]_ m[E]^=1 _ E[m/2]=m&1 _ R(*E,&,c)_(a=c B write(1,E,1)B time(j+3),memcpy(r+u(8,3,),localtime(j+3),m)),a<2?*E=~lseek(O=4[E][j],a(I)5[i]<<9,0)?(a?write:read)(O,r+u(8,3,),*i):0:0),O=u,D(16)?v(0):D(17)&&G(F(0)),CX*D(20)+D(18)-D(19)*~!!L,D(15)?O=m=N,41[43[44[E]=h(N),E]=!N,E]=D(50):0,!++q?kb=1,*l?SDL_PumpEvents(),k=k?k:SDL_SetVideoMode(720,348,32,0),DX():k?SDL_Quit(),k=0:0:0;}i(F,40[E]=!!o)i(z,42[E]=!!o)i(G,48[E]=o)
Correct.
I thought you would actually debug the heck out of that and shrug the link break error as my fault for using Android to post that.
Try this on for size.
Apple didnt spot the bug, im sure you wont be able to 
Actually I spotted that bug as soon as I saw the if statements and the first if statement didnt have brackets, when I did a course in University, our tutor kept pretty much nailing it into us to try and not use single line if statements, he said if your working on the project by your self go ahead, but in a group project, someone may have the standard of always using brackets, and could just shove some new code for the if statement in the single line if statement.
so soon as I saw the if statements, I checked and saw the second [icode]goto fail;[/icode] immediately.
but knowing I was looking for a bug may of helped me spotting it as soon as I saw it.
Also should there be [icode]uint8_t[/icode] instead of [icode]uint8 t[/icode]? It took me 5 minutes to notice that. He used the same [icode]goto[/icode] statement twice without braces on the if statement.
int ll = 2;
int swimmer = l1;
l1 = swimner;
Depends on your font.
It’s declared as [icode]ll[/icode] but used as [icode]l1[/icode].
Look closer
Here’s a fun one.
#define six 1+5
#define nine 8+1
int main()
{
printf("%d\n", six*nine);
return 0;
}
Bonus Hitchhiker’s Guide to the Galaxy reference.
May I solve this one? 
[spoiler]Since the macros “six” and “nine” in the expression “six * nine” expand to “1+5*8+1”, the result is 42, not 54 as expected. Haven’t tried it, though…[spoiler]
C preprocessor directives act even before the compulsion starts. So the result will be 42 instead of 54 which is expected.