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Showing content from https://en.cppreference.com/w/cpp/language/../error/error_code/../../numeric/math/../math/pow.html below:

std::pow, std::powf, std::powl - cppreference.com

float       powf( float base, float exp );

long double powl( long double base, long double exp );

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.

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.

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.

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)).

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.

#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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