std::fdim, std::fdimf, std::fdiml
<cmath>
double fdim ( double x, double y );
fdim ( /*floating-point-type*/ x,
(constexpr since C++23)
(constexpr since C++23)
<simd>
constexpr /*math-common-simd-t*/<V0, V1>
<cmath>
double fdim ( Integer x, Integer y );
std::fdim for all cv-unqualified floating-point types as the type of the parameters.(since C++23)std::fdim on v_xand v_y.- (See math-common-simd-t for its definition.)
[edit] Parameters
[edit] Return value
If successful, returns the positive difference between x and y.
If a range error due to overflow occurs, +HUGE_VAL , +HUGE_VALF, or +HUGE_VALL is returned.
If a range error due to underflow occurs, the correct value (after rounding) is returned.
[edit] Error handling
Errors are reported as specified in math_errhandling .
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
- If either argument is NaN, NaN is returned.
[edit] Notes
Equivalent to std::fmax (x - y, 0), except for the NaN handling requirements.
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::fdim(num1, num2) has the same effect as std::fdim(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::fdim(num1, num2) has the same effect as std::fdim(static_cast<double>(num1),
static_cast<double>(num2)). - Otherwise, if num1 or num2 has type float, then std::fdim(num1, num2) has the same effect as std::fdim(static_cast<float>(num1),
static_cast<float>(num2)).
If num1 and num2 have arithmetic types, then std::fdim(num1, num2) has the same effect as std::fdim(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 <cerrno> #include <cfenv> #include <cmath> #include <cstring> #include <iostream> #ifndef __GNUC__ #pragma STDC FENV_ACCESS ON #endif int main() { std::cout << "fdim(4, 1) = " << std::fdim(4, 1) << '\n' << "fdim(1, 4) = " << std::fdim(1, 4) << '\n' << "fdim(4,-1) = " << std::fdim(4, -1) << '\n' << "fdim(1,-4) = " << std::fdim(1, -4) << '\n'; // error handling errno = 0; std::feclearexcept (FE_ALL_EXCEPT ); std::cout << "fdim(1e308, -1e308) = " << std::fdim(1e308, -1e308) << '\n'; if (errno == ERANGE ) std::cout << " errno == ERANGE: " << std::strerror (errno) << '\n'; if (std::fetestexcept (FE_OVERFLOW )) std::cout << " FE_OVERFLOW raised\n"; }
Output:
fdim(4, 1) = 3 fdim(1, 4) = 0 fdim(4,-1) = 5 fdim(1,-4) = 5 fdim(1e308, -1e308) = inf errno == ERANGE: Numerical result out of range FE_OVERFLOW raised