ioccc/1993/vanb

To build:

    make all

Bugs and (Mis)features:

The current status of this entry is:

STATUS: INABIAF - please DO NOT fix

For more detailed information see 1993/vanb in bugs.html.

To use:

    ./vanb 'exp'

where exp is an octal expression. See below for how it can also have hexadecimal and decimal numbers.

Try:

    ./try.sh

What happens if you use something other than valid digits?

Judges' remarks:

The octal expression may contain:

• unary operators: +, -
• binary operators: +, -, *, /, %
• hex numbers: x or X followed by hex digits
• decimal numbers: d or D followed by decimal digits
• octal numbers: octal digits
• grouping: ()

No spaces are allowed in the expression. To avoid shell expansion, one should surround the expression in single quotes.

It is a good thing that this program consists of only one expression, otherwise it might become hard to understand. :-)

Author's remarks:

The program does no error checking - erroneous expressions will produce spurious results. Note that unary - is an operator. Thus, decimal -46 would be entered as -d46 and not d-46.

What Makes This Program Special:

0. Of course, the fact that the program takes input and produces output in octal, rather than a more useful system like decimal, makes it "special".

1. The entire program consists of a single expression, returned from main().

2. There are no global or local variables other than the parameters to main().

3. All of the constants in the program are expressed in octal - including the printf(3) string.

4. The variables are given names which cause visual confusion with the octal constants.

5. The program is formatted like a Roman numeral 8. The significance of 8 is obvious. I chose Roman numerals because it is a numbering system even more arcane than octal. It's not very 'deceptive', but it's difficult to deceptively format a program which has no explicit control structures (other than ?:s).

6. Many simple tasks are done many times. I tried to do these differently each time, to make the program harder to understand.

How It Works:

It's a recursive descent parser, calling main() for all of its recursion. O2 is a state variable, telling main() what grammar non-terminal to parse. **O7 is the next character. O3 is an intermediate result. Whenever you see a construct like:

    !(expression with O2)?

it's decrementing and testing O2 to see what state it's in. Comparisons involving **O7 and octal constants are looking for certain characters.

Here's the grammar ('e' denotes the empty string) :

    <E>  ::=  <T><E'>
    <E'> ::=  +<T><E'> |
              -<T><E'> | e
    <T>  ::=  <F><T'>
    <T'> ::=  *<F><T'> |
              /<F><T'> |
              %<F><T'> | e
    <F>  ::=  +<F> |
              -<F> |
              (<E><C> |
              d<D> |
              D<D> |
              x<X> |
              X<X> |
              <O>
    <C>  ::=  )
    <D>  ::=  [0-9]<D> |
              e
    <X>  ::=  [0-9]<X> |
              [A-F]<X> |
              [a-f]<X> |
              e
    <O>  ::=  [0-7]<O> |
              e

Here's how the grammar nonterminals map to octal state numbers:

    <E>  is  012
    <E'> is  011
    <T>  is  010
    <T'> is   07
    <F>  is   06
    <C>  is  013
    <D>  is   04
    <X>  is   05
    <O>  is   03

Here's a version of the program before it got formatted into the VIII, augmented with comments showing where each state begins. N1 and N2 are notes:

    #define O5 main
    O5(O2,O7,O3)char**O7;
    {
             return
    /* N1 */ !(O2+=~01+01) ?\
                  00:
    /* N2 */ !(O2-=02>01) ?\
                  printf("\045\157\012",O5(012,O7+01,00)):
    /* O  */ !(O2-=02>>01) ?\
                  (**O7<=067 && **O7>057 ? O5(03,O7,*(*O7)++-060+010*O3):O3):
    /* D  */ !(O2-=-O3-~O3) ?\
                  (072>**O7 && 060<=**O7 ? O5(04,O7,012*O3-060+*(*O7)++):O3):
    /* X  */ !(O2-=!O3+!!O3) ?\
                  (**O7>057 && **O7<=071 ? O5(05,O7,*(*O7)+++O3*020-060):
                  **O7<=0106 && 0101<=**O7 ? O5(05,O7,020*O3+*(*O7)++-067):
                  0140<**O7 && **O7<0147 ? O5(05,O7,-0127+*(*O7)+++020*O3):O3):
    /* F  */ !(O2-=02-01) ?\
                  (**O7==050 ? 050**++*O7,O5(013,O7,O5(012,O7,00)):
                  **O7<056 && 054<**O7 ? 055**++*O7,-O5(06,O7,00):
                  054>**O7 && 052<**O7 ? 050**(*O7)++,O5(06,O7,00):
                  !(**O7^0170)||!(0130^**O7) ? *++*O7,O5(05,O7,00):
                  **O7==0144||**O7==0104 ? ++*O7,O5(04,O7,00):O5(03,O7,00)):
    /* T' */ !--O2 ?\
                  (**O7==052 ? O5(07,O7,O3*(*++*O7,O5(06,O7,00))):
                  !(045-**O7) ? O5(07,O7,O3%(03+(*O7)++,O5(06,O7,00))):
                  !(**O7^057) ? O5(07,O7,O3/(03-*++*O7,O5(06,O7,00))):O3):
    /* T  */ !(O2+=01-02) ?\
                  O5(07,O7,O5(06,O7,00)):
    /* E' */ !(O2+=-02/02) ?\
                  (!(**O7-053)?O5(011,O7,O3+(++*O7,O5(010,O7,00))):
                  !(055^**O7) ? O5(011,O7,O3-(03+*(*O7)++,O5(010,O7,00))):O3):
    /* E  */ !(O2-=0563&0215) ?\
                  O5(011,O7,O5(010,O7,00)):
    /* C  */ (++*O7,O3);
    }

Note N1: It should never enter this state, unless the user invokes the program with no parameters, in which case it just returns 0.

Note N2: Since the program is properly invoked with 1 parameter, this is the first state it will go into. This state just invokes printf, and sends the parser to state 012 (which is <E>).

The E and T states work like this:

    int e(){ return eprime( t() ); }

The E' and T' states work like this:

    int eprime( int intermediate )
    {
        if( ch == '+' )
        {
            ch = nextchar();
            return eprime( intermediate + t() );
        }
        else return intermediate;
    }

The D, X and O states work like this (they assume that they're initially called with 0):

    int octal( int intermediate )
    {
        if( ch>='0' && ch<='7' )
        {
            intermediate = intermediate * 8 + ch - '0';
            ch = nextchar();
            return octal( intermediate );
        }
        else return intermediate;
    }

F and C work similarly.