std::pow, std::powf, std::powl

Defined in header `<cmath>`
	(1)
float pow ( float base, float exp ); double pow ( double base, double exp ); long double pow ( long double base, long double exp );			(until C++23)
/* floating-point-type / pow ( / floating-point-type / base, / floating-point-type */ exp )			(since C++23) (constexpr since C++26)
float pow ( float base, int exp ); double pow ( double base, int exp ); long double pow ( long 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)
Defined in header `<cmath>`
template< class Arithmetic1, class Arithmetic2 > /* common-floating-point-type */ pow ( Arithmetic1 base, Arithmetic2 exp );		(A)	(constexpr since C++26)

1-4) Computes the value of base raised to the power exp. The library provides overloads of 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.

(since C++11)

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:

If num1 or num2 has type long double, then std::pow(num1, num2) has the same effect as std::pow(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::pow(num1, num2) has the same effect as std::pow(static_cast<double>(num1),
static_cast<double>(num2)).
Otherwise, if num1 or num2 has type float, then std::pow(num1, num2) has the same effect as std::pow(static_cast<float>(num1),
static_cast<float>(num2)).

(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),
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)

Example

Run this code

#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

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External libraries

sqrtsqrtfsqrtl (C++11)(C++11)	computes square root (\(\small{\sqrt{x}}\) $\sqrt x$ ) (function) [edit]
cbrtcbrtfcbrtl (C++11)(C++11)(C++11)	computes cube root (\(\small{\sqrt[3]{x}}\) $3 \sqrt x$ ) (function) [edit]
hypothypotfhypotl (C++11)(C++11)(C++11)	computes square root of the sum of the squares of two or three(since C++17) given numbers (\(\scriptsize{\sqrt{x^2+y^2}}\) $\sqrt x 2 +y 2$ ), (\(\scriptsize{\sqrt{x^2+y^2+z^2}}\) $\sqrt x 2 +y 2 +z 2$ )(since C++17) (function) [edit]
pow(std::complex)	complex power, one or both arguments may be a complex number (function template) [edit]
pow(std::valarray)	applies the function std::pow to two valarrays or a valarray and a value (function template) [edit]
C documentation for pow

cppreference.com

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std::pow, std::powf, std::powl

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Example

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base	-	base as a floating-point or integer value
exp	-	exponent as a floating-point or integer value