std::cbrt, std::cbrtf, std::cbrtl
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Defined in header
<cmath>
(1)
float cbrt ( float num );
(until C++23)
double cbrt ( double num );
/*floating-point-type*/
cbrt ( /*floating-point-type*/ num );
(since C++23) cbrt ( /*floating-point-type*/ num );
(constexpr since C++26)
float cbrtf( float num );
(2)
(since C++11) (constexpr since C++26)
long double cbrtl( long double num );
(3)
(since C++11) (constexpr since C++26)
SIMD overload (since C++26)
Defined in header
<simd>
template< /*math-floating-point*/ V >
(S)
(since C++26)
constexpr /*deduced-simd-t*/<V>
Additional overloads (since C++11)
Defined in header
<cmath>
template< class Integer >
double cbrt ( Integer num );
(A)
(constexpr since C++26)
double cbrt ( Integer num );
1-3) Computes the cube root of num. The library provides overloads of
std::cbrt for all cv-unqualified floating-point types as the type of the parameter.(since C++23)S) The SIMD overload performs an element-wise
std::cbrt on v_num.- (See math-floating-point and deduced-simd-t for their definitions.)
A) Additional overloads are provided for all integer types, which are treated as double.
(since C++11)[edit] Parameters
num
-
floating-point or integer value
[edit] Return value
If no errors occur, the cube root of num (\(\small{\sqrt[3]{num} }\)3√num), is returned.
If a range error occurs due to underflow, the correct result (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 the argument is ±0 or ±∞, it is returned, unchanged.
- if the argument is NaN, NaN is returned.
[edit] Notes
std::cbrt(num) is not equivalent to std::pow (num, 1.0 / 3) because the rational number \(\small{\frac1{3} }\) 1
3
is typically not equal to 1.0 / 3 and std::pow cannot raise a negative base to a fractional exponent. Moreover, std::cbrt(num) usually gives more accurate results than std::pow (num, 1.0 / 3) (see example).
The additional overloads are not required to be provided exactly as (A). They only need to be sufficient to ensure that for their argument num of integer type, std::cbrt(num) has the same effect as std::cbrt(static_cast<double>(num)).
[edit] Example
Run this code
#include <cmath> #include <iomanip> #include <iostream> #include <limits> int main() { std::cout << "Normal use:\n" << "cbrt(729) = " << std::cbrt(729) << '\n' << "cbrt(-0.125) = " << std::cbrt(-0.125) << '\n' << "Special values:\n" << "cbrt(-0) = " << std::cbrt(-0.0) << '\n' << "cbrt(+inf) = " << std::cbrt(INFINITY ) << '\n' << "Accuracy and comparison with `pow`:\n" << std::setprecision (std::numeric_limits <double>::max_digits10) << "cbrt(343) = " << std::cbrt(343) << '\n' << "pow(343,1.0/3) = " << std::pow (343, 1.0 / 3) << '\n' << "cbrt(-343) = " << std::cbrt(-343) << '\n' << "pow(-343,1.0/3) = " << std::pow (-343, 1.0 / 3) << '\n'; }
Possible output:
Normal use: cbrt(729) = 9 cbrt(-0.125) = -0.5 Special values: cbrt(-0) = -0 cbrt(+inf) = inf Accuracy and comparison with `pow`: cbrt(343) = 7 pow(343,1.0/3) = 6.9999999999999991 cbrt(-343) = -7 pow(-343,1.0/3) = -nan
[edit] See also
(C++11)(C++11)(C++11)
+y2
and \(\scriptsize{\sqrt{x^2+y^2+z^2}}\)√x2
+y2
+z2
(since C++17)
(function) [edit]
C documentation for cbrt