What are the actual min/max values for float and double (C++)
The std::numerics_limits
class in the <limits>
header provides information about the characteristics of numeric types.
For a floating-point type T
, here are the greatest and least values representable in the type, in various senses of “greatest” and “least.” I also include the values for the common IEEE 754 64-bit binary type, which is called double
in this answer. These are in decreasing order:
std::numeric_limits<T>::infinity()
is the largest representable value, ifT
supports infinity. It is, of course, infinity. Whether the typeT
supports infinity is indicated bystd::numeric_limits<T>::has_infinity
.std::numeric_limits<T>::max()
is the largest finite value. Fordouble
, this is 21024−2971, approximately 1.79769•10308.std::numeric_limits<T>::min()
is the smallest positive normal value. Floating-point formats often have an interval where the exponent cannot get any smaller, but the significand (fraction portion of the number) is allowed to get smaller until it reaches zero. This comes at the expense of precision but has some desirable mathematical-computing properties.min()
is the point where this precision loss starts. Fordouble
, this is 2−1022, approximately 2.22507•10−308.std::numeric_limits<T>::denorm_min()
is the smallest positive value. In types which have subnormal values, it is subnormal. Otherwise, it equalsstd::numeric_limits<T>::min()
. Fordouble
, this is 2−1074, approximately 4.94066•10−324.std::numeric_limits<T>::lowest()
is the least finite value. It is usually a negative number large in magnitude. Fordouble
, this is −(21024−2971), approximately −1.79769•10308.If
std::numeric_limits<T>::has_infinity
andstd::numeric_limits<T>::is_signed
are true, then-std::numeric_limits<T>::infinity()
is the least value. It is, of course, negative infinity.
Another characteristic you may be interested in is:
std::numeric_limits<T>::digits10
is the greatest number of decimal digits such that converting any decimal number with that many digits toT
and then converting back to the same number of decimal digits will yield the original number. Fordouble
, this is 15.
It's all to be found in numeric_limits.
But BEWARE
For some reason unknown to me, std::numeric_limits<float>:min()
does not return the minimum float. Instead it returns the smallest positive float that is represented in normalized form. To get the minimum, use std::numeric_limits<float>::lowest()
. I kid you not. Likewise for other floating point types, i.e. double
and long double
.
http://en.cppreference.com/w/cpp/types/numeric_limits
Alright. Using what I learned from here (thanks everyone) and the other parts of the web I wrote a neat little summary of the two just in case I run into another issue like this.
In C++ there are two ways to represent/store decimal values.
Floats and Doubles
A float can store values from:
- -340282346638528859811704183484516925440.0000000000000000 Float lowest
- 340282346638528859811704183484516925440.0000000000000000 Float max
A double can store values from:
-179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368.0000000000000000 Double lowest
179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368.0000000000000000 Double max
Float's precision allows it to store a value of up to 9 digits (7 real digits, +2 from decimal to binary conversion)
Double, like the name suggests can store twice as much precision as a float. It can store up to 17 digits. (15 real digits, +2 from decimal to binary conversion)
e.g.
float x = 1.426;
double y = 8.739437;
Decimals & Math
Due to a float being able to carry 7 real decimals, and a double being able to carry 15 real decimals, to print them out when performing calculations a proper method must be used.
e.g
include
typedef std::numeric_limits<double> dbl;
cout.precision(dbl::max_digits10-2); // sets the precision to the *proper* amount of digits.
cout << dbl::max_digits10 <<endl; // prints 17.
double x = 12345678.312;
double a = 12345678.244;
// these calculations won't perform correctly be printed correctly without setting the precision.
cout << endl << x+a <<endl;
example 2:
typedef std::numeric_limits< float> flt;
cout.precision(flt::max_digits10-2);
cout << flt::max_digits10 <<endl;
float x = 54.122111;
float a = 11.323111;
cout << endl << x+a <<endl; /* without setting precison this outputs a different value, as well as making sure we're *limited* to 7 digits. If we were to enter another digit before the decimal point, the digits on the right would be one less, as there can only be 7. Doubles work in the same way */
Roughly how accurate is this description? Can it be used as a standard when confused?