Bug in boost/rational.hpp

From: John (EBo) David (ebo_at_[hidden])
Date: 2000-11-18 23:10:17

If the following bug has already been reported my apologies...

Description and reproducion:

  The devision of a rational by "0" does not throw an exception but
  returns a 0 denominator in the result.  If you trace through the
  rational code example you will find that the exception is actually
  never generated by the test code.

Effected code segment:

  boost/rational.hpp: 195-206

Proposed Fix:

  Add a bad_rational test (denominator == 0) test and throw an
  exception.

  template <typename IntType>
  rational<IntType>& rational<IntType>::operator/= (const
rational<IntType>& r)
  {
    // Avoid overflow and preserve normalization
    IntType gcd1 = gcd<IntType>(num, r.num);
    IntType gcd2 = gcd<IntType>(r.den, den);
    num = (num/gcd1) * (r.den/gcd2);
    den = (den/gcd2) * (r.num/gcd1);
>>>    // validity check
>>>    if (0 == den) throw bad_rational();
    return *this;
  }

Notes:

  as an aside, years ago I learned a little programming trick which I
  used above.  By placing the literal on the left hand side of the
  expression I use compilers parser to catch certain types of typos.
  In this way if I inadvertently type a "=" instead of a "==" the
  compiler will see it as a syntax error.  i.e. a "if (0=den)"
  produces a syntax error where as "if (den=0)" is valid but not what
  I wanted.  Just an aside...