std::fmax, std::fmaxf, std::fmaxl
<cmath>
double fmax ( double x, double y );
fmax ( /*floating-point-type*/ x,
(constexpr since C++23)
(constexpr since C++23)
<simd>
constexpr /*math-common-simd-t*/<V0, V1>
<cmath>
double fmax ( Integer x, Integer y );
std::fmax
for all cv-unqualified floating-point types as the type of the parameters.(since C++23)std::fmax
on v_xand v_y.- (See math-common-simd-t for its definition.)
[edit] Parameters
[edit] 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.
[edit] 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.
[edit] 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)).
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
Output:
fmax(2,1) = 2 fmax(-Inf,0) = 0 fmax(NaN,-1) = -1