The shortest way to convert infix expressions to postfix (RPN) in C

Question

Original formulation is given here (you can try also your program for correctness) .

Additional rules:
1. The program should read from standard input and write to s

Answer 1

Here's a way to reduce the code down to 76 chars:

main(c){read(0,&c,1)?c-41&&main(c-40&&putchar(c,c%96>26&&main(c))):exit(0);}

A longer, commented version for clarity:

int main(int c)
{
   if (read(0,&c,1)) {          /* read char */
       if (c-41) {              /* if not ')' */
           if (c-40) {          /* then if not '(' */
               if (c%96>26) {   /* then if operator (not alphabet or <LF>) */
                   main(c);     /* recurse */
               }
               putchar(c);      /* print */
           }
           main(c);             /* recurse */
       }        
   } else exit(0);              /* end program */
}

Answer 2

I'm not out to break any records but I'll post this anyway:

#define x(z) while(p>##z s)putchar(*p--);
main(c){
int s[9],*p=s-1;
for(;read(0,&c,1);){
isalpha(c)?putchar(c):c=='('?(c=0):c==')'?(c=1):isdigit(c)?:(*++p=c);
if(c==0){x()main(0);}
if(c==1) break;}
x(=)return 0;}

Edit: Fixed correctness issues pointed out by kuszi's comment.

Answer 3

Well, the real winner is the one who wrote this small code you provided, but you can slightly modify it to remove the exit:

main(c){read(0,&c,1)?c-41&&main(c-40&&(c%96<27||main(c),putchar(c))):0;}

I tried and it works.