double fmax ( double x, double y );
fmax ( /*floating-point-type*/ x,
float fmaxf( float x, float y );
(2) (since C++11)long double fmaxl( long double x, long double y );
(3) (since C++11)constexpr /*math-common-simd-t*/<V0, V1>
template< class Integer >
double fmax ( Integer x, Integer y );
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)
S) The SIMD overload performs an element-wise std::fmax
on v_xand v_y.
A) Additional overloads are provided for all integer types, which are treated as double.
(since C++11) [edit] Parameters x, y - floating-point or integer values [edit] Return valueIf successful, returns the larger of two floating point values. The value returned is exact and does not depend on any rounding modes.
[edit] Error handlingThis function is not subject to any of the error conditions specified in math_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
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.
(since C++23) [edit] 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'; }
Output:
fmax(2,1) = 2 fmax(-Inf,0) = 0 fmax(NaN,-1) = -1[edit] See also checks if the first floating-point argument is greater than the second
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