Fmin
Defined in header <cmath>
.
Description
Returns the smaller of two floating point arguments, treating NaNs as missing data (between a NaN and a numeric value, the numeric value is chosen). The library provides overloads of std::fmin for all cv-unqualified floating-point types as the type of the parameters x and y.
Declarations
- C++23
- C++11
// 1)
constexpr /* floating-point-type */
fmin ( /* floating-point-type */ x,
/* floating-point-type */ y );
// 2)
constexpr float fminf( float x, float y );
// 3)
constexpr long double fminl( long double x, long double y );
// 4)
template< class Arithmetic1, class Arithmetic2 >
constexpr /* common-floating-point-type */ fmin( Arithmetic1 x, Arithmetic2 y );
// 1)
float fmin ( float x, float y );
// 2)
double fmin ( double x, double y );
// 3)
long double fmin ( long double x, long double y );
// 4)
float fminf( float x, float y );
// 5)
long double fminl( long double x, long double y );
// 6)
template< class Arithmetic1, class Arithmetic2 >
/* common-floating-point-type */ fmin( Arithmetic1 x, Arithmetic2 y );
Parameters
x
,y
- floating-point or integer values
Return Value
If successful, returns the smaller of two floating point values. The value returned is exact and does not depend on any rounding modes.
Error handling
This function is not subject to any of the error conditions specified in math_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC 60559):
If one of the two arguments is NaN, the value of the other argument is returned Only if both arguments are NaN, NaN is returned
Notes
This function is not required to be sensitive to the sign of zero,
although some implementations additionally enforce that if one argument is +0
and the other is -0
, then -0
is returned.
The additional overloads are not required to be provided exactly as Additional Overloads.
They only need to be sufficient to ensure that for their first argument num1
and second argument num2
:
If num1
or num2
has type long double, then
std::fmin(num1, num2)
has the same effect as
std::fmin(static_cast<long double>(num1), static_cast<long double>(num2))
.
Otherwise, if num1
and/or num2
has type double or an integer type, then
std::fmin(num1, num2)
has the same effect as
std::fmin(static_cast<double>(num1), static_cast<double>(num2))
.
Otherwise, if num1
or num2
has type float, then
std::fmin(num1, num2)
has the same effect as
std::fmin(static_cast<float>(num1), static_cast<float>(num2))
. (do C++23)
If num1
and num2
have arithmetic types, then
std::fmin(num1, num2)
has the same effect as
std::fmin(static_cast</* common-floating-point-type */>(num1), static_cast</* common-floating-point-type */>(num2))
where /* common-floating-point-type */ is the floating-point type with the greatest floating-point conversion rank and greatest
floating-point conversion subrank between the types of num1
and num2
, arguments of integer type are considered to have the same floating-point conversion rank as double.
If no such floating-point type with the greatest rank and subrank exists, then overload resolution does not result in a usable candidate from the overloads provided.
Examples
#include <cmath>
#include <iostream>
int main()
{
std::cout
<< "fmin(2,1) = "
<< std::fmin(2, 1) << '\n'
<< "fmin(-Inf,0) = "
<< std::fmin(-INFINITY, 0) << '\n'
<< "fmin(NaN,-1) = "
<< std::fmin(NAN, -1) << '\n';
}
fmin(2,1) = 1
fmin(-Inf,0) = -inf
fmin(NaN,-1) = -1