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From: Gabriel Dos Reis (Gabriel.Dos-Reis_at_[hidden])
Date: 2000-10-19 15:54:27

"David Abrahams" <abrahams_at_[hidden]> writes:

| How can we determine the minimum representable value of a built-in type?
|
| 18.2.1.2 numeric_limits members
| static T min() throw();
| 1 Minimum finite value. 181)
| 2 For floating types with denormalization, returns the minimum positive
| normalized value.
| 3 Meaningful for all specializations in which is_bounded != false, or
| is_bounded == false && is_signed == false.
|
| Is it guaranteed that the minimum representable value for floating types
| with denormalization is -numeric_limits<T>::max()?

Yes. As a general rule, floating point types are symmetric
(i.e. according to LIA).

| If not, I think we have a defect.
|
| If so, the rules are complicated, but it should be possible to follow them.

Actually, the rules are simple and they should not be hard to follow..

| To follow rule #2, we need to determine if the type has denormalization.
| But:
|
| 18.2.1.2 numeric_limits members 18 Language support library
| static const float_denorm_style has_denorm;
| 40 denorm_present if the type allows denormalized values (variable number of
| exponent bits) 195) , denorm_absent if the type does not allow denormalized
| values, and denorm_indeterminate if it is indeterminate at compile time
| whether the type allows denormalized values.
|
| What if has_denorm is denorm_indeterminate? Maybe the answer is:
|
| numeric_limits<T>::is_signed && numeric_limits<T>::min() >= 0
| ? -numeric_limits<T>::max() : numeric_limits<T>::min()

If the system is conforming to LIA then the floating point values set is
symmetric because LIA's model is sign-magnitude.

For non-LIA system (actually, only systems using radic-complement
representation should exhibit that behaviour) there is a notion of
most-negative value which is quite distinct from what one calls min()
in case of floating point values.

Yes, the terminology is quite confusing, but I weren't there when
LIA-1 was adopted as an international standard.

-- Gaby