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

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fmax
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Defined in header <cmath>
(1)
float       fmax ( float x, float y );

double      fmax ( double x, double y );

long double fmax ( long double x, long double y );
 (until 哋它亢++23)
constexpr /* floating-point-type */

            fmax ( /* floating-point-type */ x,

                   /* floating-point-type */ y );
 (since 哋它亢++23)
float       fmaxf( float x, float y );
 (2) (since 哋它亢++11)
(constexpr since 哋它亢++23)
long double fmaxl( long double x, long double y );
 (3) (since 哋它亢++11)
(constexpr since 哋它亢++23)
Additional overloads (since 哋它亢++11)
Defined in header <cmath>
template< class Integer >
double      fmax ( Integer x, Integer y );
 (A) (constexpr since 哋它亢++23)
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 哋它亢++23)
A) Additional overloads are provided for all integer types, which are treated as double.
 (since 哋它亢++11)

Parameters

x, y - floating-point or integer values

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.

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 (A). 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::fmax(num1, num2) has the same effect as std::fmax(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::fmax(num1, num2) has the same effect as std::fmax(static_cast<double>(num1),
              static_cast<double>(num2))    .
  • Otherwise, if num1 or num2 has type float, then std::fmax(num1, num2) has the same effect as std::fmax(static_cast<float>(num1),
              static_cast<float>(num2))    .
(until 哋它亢++23)

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.

(since 哋它亢++23)

Example

Run this code
#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';
}

Output: 

fmax(2,1)    = 2
fmax(-Inf,0) = 0
fmax(NaN,-1) = -1

See also

(哋它亢++11)
 checks if the first floating-point argument is greater than the second
(function)
(哋它亢++11)(哋它亢++11)(哋它亢++11)
 smaller of two floating point values
(function)
returns the greater of the given values
(function template)
returns the largest element in a range
(function template)
(哋它亢++11)
 returns the smaller and larger of two elements
(function template)
(哋它亢++11)
 returns the smallest and the largest elements in a range
(function template)
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