std::fmax, std::fmaxf, std::fmaxl

Since C++23

Header: <cmath>

1-3) Returns the larger 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::fmax for all cv-unqualified floating-point types as the type of the parameters.(since C++23)

# Declarations

float fmax ( float x, float y );
double fmax ( double x, double y );
long double fmax ( long double x, long double y );

(until C++23)

constexpr /*floating-point-type*/
fmax ( /*floating-point-type*/ x,
/*floating-point-type*/ y );

(since C++23)

float fmaxf( float x, float y );

(since C++11) (constexpr since C++23)

long double fmaxl( long double x, long double y );

(since C++11) (constexpr since C++23)

SIMD overload (since C++26)

template< class V0, class V1 >
constexpr /*math-common-simd-t*/<V0, V1>
fmax ( const V0& v_x, const V1& v_y );

(since C++26)

Additional overloads (since C++11)

template< class Integer >
double fmax ( Integer x, Integer y );

(constexpr since C++23)

# Parameters

# Return value

If successful, returns the larger of two floating point values. The value returned is exact and does not depend on any rounding modes.

# 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 (A). They only need to be sufficient to ensure that for their first argument num1 and second argument num2:

If num1 and num2 have arithmetic types, then std::fmax(num1, num2) has the same effect as std::fmax(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.

# Example

#include <cmath>
#include <iostream>
 
int main()
{
    std::cout << "fmax(2,1)    = " << std::fmax(2, 1) << '\n'
              << "fmax(-Inf,0) = " << std::fmax(-INFINITY, 0) << '\n'
              << "fmax(NaN,-1) = " << std::fmax(NAN, -1) << '\n';
}

# See also