CODEWITHSUNDEEP
c++c++11
const_ref
const_ref

Reputation: 4096

Issue with std::numeric_limits<T>::min() with float/double values

While writing some code to ensure user input is valid I came across an issue with std::numeric_limits<T>::min(); where T is a floating point type i.e double/float etc.

Using -std::numeric_limits<double>::max(); allows me to get the real minimum value but you would expect std::numeric_limits<double>::min(); to return that.

Why does std::numeric_limits<double>::min(); not return the smallest possible value of these types and instead force us to use std::numeric_limits::lowest or -std::numeric_limits<double>::max();?

Upvotes: 2

Views: 2608

Answers (2)

4pie0
4pie0

Reputation: 29724

Why is this?

This is because std::numeric_limits<>::min() returns implementation defined FLT_MIN, DBL_MIN or INT_MIN. In this regard behavior of this method is consistent. But the return value ofstd::numeric_limits<double>::min(): DBL_MIN has slightly different meaning than INT_MIN. This is the smallest value that double can represent. There might be value greater than 0 but double can't represent it.

why does using min() not do the same thing in this instance?

The rationale behind this is that you can query <limits> for this value to check for this possible underflow specific to floating point arithmetic.

You can use

std::numeric_limits<double>::lowest()

to query for a lowest negative value.

Upvotes: 0

Anycorn
Anycorn

Reputation: 51465

Yes, using -max() will give lowest value.

In C++11 there is also std::numeric_limits::lowest that is consistent for reals and ints. http://en.cppreference.com/w/cpp/types/numeric_limits/lowest

See also How to workaround the inconsistent definition of numeric_limits<T>::min()?

Upvotes: 1

Related Questions