std::pow, std::powf, std::powl
Defined in header <cmath>
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(1) | ||
float pow ( float base, float exp ); double pow ( double base, double exp ); |
(until C++23) | |
/* floating-point-type */ pow ( /* floating-point-type */ base, |
(since C++23) (constexpr since C++26) |
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float pow ( float base, int exp ); double pow ( double base, int exp ); |
(2) | (until C++11) |
float powf( float base, float exp ); |
(3) | (since C++11) (constexpr since C++26) |
long double powl( long double base, long double exp ); |
(4) | (since C++11) (constexpr since C++26) |
Additional overloads (since C++11) |
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Defined in header <cmath>
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template< class Arithmetic1, class Arithmetic2 > /* common-floating-point-type */ |
(A) | (constexpr since C++26) |
std::pow
for all cv-unqualified floating-point types as the type of the parameters base and exp. (since C++23)
A) Additional overloads are provided for all other combinations of arithmetic types.
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(since C++11) |
Parameters
base | - | base as a floating-point or integer value |
exp | - | exponent as a floating-point or integer value |
Return value
If no errors occur, base raised to the power of exp (baseexp
), is returned.
If a domain error occurs, an implementation-defined value is returned (NaN where supported).
If a pole error or a range error due to overflow occurs, ±HUGE_VAL, ±HUGE_VALF
, or ±HUGE_VALL
is returned.
If a range error occurs due to underflow, the correct result (after rounding) is returned.
Error handling
Errors are reported as specified in math_errhandling.
If base is finite and negative and exp is finite and non-integer, a domain error occurs and a range error may occur.
If base is zero and exp is zero, a domain error may occur.
If base is zero and exp is negative, a domain error or a pole error may occur.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
- pow(+0, exp), where exp is a negative odd integer, returns +∞ and raises FE_DIVBYZERO
- pow(-0, exp), where exp is a negative odd integer, returns -∞ and raises FE_DIVBYZERO
- pow(±0, exp), where exp is negative, finite, and is an even integer or a non-integer, returns +∞ and raises FE_DIVBYZERO
- pow(±0, -∞) returns +∞ and may raise FE_DIVBYZERO
- pow(+0, exp), where exp is a positive odd integer, returns +0
- pow(-0, exp), where exp is a positive odd integer, returns -0
- pow(±0, exp), where exp is positive non-integer or a positive even integer, returns +0
- pow(-1, ±∞) returns 1
- pow(+1, exp) returns 1 for any exp, even when exp is NaN
- pow(base, ±0) returns 1 for any base, even when base is NaN
- pow(base, exp) returns NaN and raises FE_INVALID if base is finite and negative and exp is finite and non-integer.
- pow(base, -∞) returns +∞ for any
|base|<1
- pow(base, -∞) returns +0 for any
|base|>1
- pow(base, +∞) returns +0 for any
|base|<1
- pow(base, +∞) returns +∞ for any
|base|>1
- pow(-∞, exp) returns -0 if exp is a negative odd integer
- pow(-∞, exp) returns +0 if exp is a negative non-integer or negative even integer
- pow(-∞, exp) returns -∞ if exp is a positive odd integer
- pow(-∞, exp) returns +∞ if exp is a positive non-integer or positive even integer
- pow(+∞, exp) returns +0 for any negative exp
- pow(+∞, exp) returns +∞ for any positive exp
- except where specified above, if any argument is NaN, NaN is returned
Notes
C++98 added overloads where exp has type int on top of C pow(), and the return type of std::pow(float, int) was float. However, the additional overloads introduced in C++11 specify that std::pow(float, int) should return double. LWG issue 550 was raised to target this conflict, and the resolution is to removed the extra int exp overloads.
Although std::pow
cannot be used to obtain a root of a negative number, std::cbrt is provided for the common case where exp is 1/3.
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:
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(until C++23) |
If num1 and num2 have arithmetic types, then std::pow(num1, num2) has the same effect as std::pow(static_cast</* common-floating-point-type */>(num1), 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) |
Example
#include <cerrno> #include <cfenv> #include <cmath> #include <cstring> #include <iostream> // #pragma STDC FENV_ACCESS ON int main() { // typical usage std::cout << "pow(2, 10) = " << std::pow(2, 10) << '\n' << "pow(2, 0.5) = " << std::pow(2, 0.5) << '\n' << "pow(-2, -3) = " << std::pow(-2, -3) << '\n'; // special values std::cout << "pow(-1, NAN) = " << std::pow(-1, NAN) << '\n' << "pow(+1, NAN) = " << std::pow(+1, NAN) << '\n' << "pow(INFINITY, 2) = " << std::pow(INFINITY, 2) << '\n' << "pow(INFINITY, -1) = " << std::pow(INFINITY, -1) << '\n'; // error handling errno = 0; std::feclearexcept(FE_ALL_EXCEPT); std::cout << "pow(-1, 1/3) = " << std::pow(-1, 1.0 / 3) << '\n'; if (errno == EDOM) std::cout << " errno == EDOM " << std::strerror(errno) << '\n'; if (std::fetestexcept(FE_INVALID)) std::cout << " FE_INVALID raised\n"; std::feclearexcept(FE_ALL_EXCEPT); std::cout << "pow(-0, -3) = " << std::pow(-0.0, -3) << '\n'; if (std::fetestexcept(FE_DIVBYZERO)) std::cout << " FE_DIVBYZERO raised\n"; }
Possible output:
pow(2, 10) = 1024 pow(2, 0.5) = 1.41421 pow(-2, -3) = -0.125 pow(-1, NAN) = nan pow(+1, NAN) = 1 pow(INFINITY, 2) = inf pow(INFINITY, -1) = 0 pow(-1, 1/3) = -nan errno == EDOM Numerical argument out of domain FE_INVALID raised pow(-0, -3) = -inf FE_DIVBYZERO raised
See also
(C++11)(C++11) |
computes square root (√x) (function) |
(C++11)(C++11)(C++11) |
computes cube root (3√x) (function) |
(C++11)(C++11)(C++11) |
computes square root of the sum of the squares of two or three (since C++17) given numbers (√x2 +y2 ), (√x2 +y2 +z2 ) (since C++17) (function) |
complex power, one or both arguments may be a complex number (function template) | |
applies the function std::pow to two valarrays or a valarray and a value (function template) |