Re: what does numeric_limits<double>::min() mean?
In article <PKUCj.9990$%15.6841@bignews7.bellsouth.net>, Edward Diener
<diener896092_no_spam_here@bellsouth.net> wrote:
vact.lym@gmail.com wrote:
I think numeric_limits<double>::min() is the smallest value which can
be stored in a double type variable, but cout <<
numeric_limits<double>::min() / 2; gives a non-zero value.
cout << numeric_limits<double>::min() / 100; is non-zero too.
so what does numeric_limits<double>::min() mean?
thank you.
The smallest value is a negative number for signed types. A double is a
signed type.
the truth is
<quote>
18.2.1.2 numeric_limits members [numeric.limits.members]
static constexpr T min() throw();
Minimum finite value.189)
For floating types with denormalization, returns the minimum positive
normalized value.
Meaningful for all specializations in which is_bounded != false, or
is_bounded == false && is_signed
== false.
<\quote>
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