0 
See also
- float/1 to convert to float.
- Wikipedia: Double-precision floating-point format
For fixed-point arithmetic (e.g. involving monetary amounts), "rational numbers" (provided by the GMP library) are the correct choice:
Floats of note
Negative infinity
?- NegInf is -1.0Inf, float(NegInf). NegInf = -1.0Inf.
Positive infinity
?- PosInf is +1.0Inf, float(PosInf). PosInf = 1.0Inf.
?- What is +1.0Inf + -1.0Inf. ERROR: Arithmetic: evaluation error: `undefined'
Not-a-Number
?- NaN is nan, float(NaN). NaN = 1.5NaN.
Machine epsilon (the smallest one, which, added to 1.0, still gives a result larger than 1.0):
?- Macheps is epsilon, float(Macheps). Macheps = 2.220446049250313e-16.
Negative Zero
?- NegZero is -0.0, float(NegZero), NegZero =:= 0.0, format("~g",[NegZero]).
-0
MinusZero = -0.0.
There is no function to get the smallest-nonzero/largest floats though.
What is a "float"?
According to the Prolog ISO standard (ISO/IEC 13211), chapter 7.1.3, a "floating point" value is a "a member of a set F, which is a finite subset of R" (should really be Q, right?) for which a floating-point representation exists (basically) with the standard using what's in ISO/IEC 10967, Language independent arithmetic (LIA).
There is no advocation in the standard for IEEE 754 Floating-Points but practice that is what will be used as that is what:
- is compatible with ISO/IEC 10967 (apparently)
- is what the hardware supports (which leads to interesting questions about NaNs, +0, -0, +oo, -oo and floating point operation error flags)
See:
- Notes on SWI-Prolog's IEEE 754 floating point arithmetic
- Note on ISO Prolog conflicting with IEEE 754
Assuming we are using 64-bit double precision IEEE floating points, then:
(The internal representation of a float uses base 2, so a base 10 is approximate)
Largest float:
?- X=1.7976931348623157e308, Y is X+1, Y=X.
Smallest normalized float:
X = 2.2250738585072014e-308
Smallest unnormalized float:
?- X=4.9406564584124654e-324, Y is X/2.0.
Machine epsilon (the smallest one, which, added to 1.0, still gives a result larger than 1.0):
efloat(C,E) :-
erecur(C,1.0,E).
erecur(C,X,E) :-
Xhalf is X / 2.0,
Xtest is C + Xhalf,
Xtest \= C,
!,
format("~e/2.0 = ~20f\n",[Xhalf,Xtest]),
erecur(Xhalf,E).
erecur(C,X,X) :-
Xhalf is X / 2.0,
Xtest is C + Xhalf,
Xtest = C,
format("~e",[X]).
Then:
efloat(1.0,X). X = 2.220446049250313e-16.
or one can ask SWI-Prolog directly:
?- X is epsilon. X = 2.220446049250313e-16.