Question

OK, so I know you’re generally not supposed to compare two floating-point numbers for equality. However, in William Kahan’s How Futile are Mindless Assessments of Roundoff in Floating-Point Computation? he shows the following code (pseudo-code, I believe):

Real Function T(Real z) :
      T := exp(z) ;                       ... rounded, of course.
      If (T = 1) Return( T ) ;            ... when |z| is very tiny.
      If (T = 0) Return( T := –1/z ) ;    ... when exp(z) underflows.
      Return( T := ( T – 1 )/log(T) ) ;   ... in all other cases.
      End T .

Now, I’m interested in implementing this in C or C++, and I have two related questions:

a) if I take T to be a double, then in the comparison (T == 1) or (T == 0) would 0 and 1 get converted to double to preserve the precision of the values involved in a multi-type expression?

b) does this still count as comparing two floating-point numbers for equality?

Answer 1

Yes and yes.

For 32-bit ints, double can represent every value precisely. When you compare a double to a 64-bit int, however, there will be potential roundoff error if the int is greater than 2^52. You can use long double, though, which has at least 64 bits of mantissa.

Of course, the best way is just to use a floating-point literal: 1.0 or just 1. has type double, 1.0f is a float, and my_float_type(1) has whatever type it’s supposed to :v) .