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

std::ldexp, std::ldexpf, std::ldexpl - cppreference.com

constexpr /* floating-point-type */
            ldexp ( /* floating-point-type */ num, int exp );

float       ldexpf( float num, int exp );

long double ldexpl( long double num, int exp );

template< class Integer >
double      ldexp ( Integer num, int exp );

1-3) Multiplies a floating point value num by the number 2 raised to the exp power. The library provides overloads of std::ldexp for all cv-unqualified floating-point types as the type of the parameter num.(since C++23)

A) Additional overloads are provided for all integer types, which are treated as double.

If no errors occur, num multiplied by 2 to the power of exp (num×2exp
) is returned.

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 result (after rounding) is returned.

Errors are reported as specified in math_errhandling.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

• Unless a range error occurs, FE_INEXACT is never raised (the result is exact).
• Unless a range error occurs, the current rounding mode is ignored.
• If num is ±0, it is returned, unmodified.
• If num is ±∞, it is returned, unmodified.
• If exp is 0, then num is returned, unmodified.
• If num is NaN, NaN is returned.

On binary systems (where FLT_RADIX is 2), std::ldexp is equivalent to std::scalbn.

The function std::ldexp ("load exponent"), together with its dual, std::frexp, can be used to manipulate the representation of a floating-point number without direct bit manipulations.

On many implementations, std::ldexp is less efficient than multiplication or division by a power of two using arithmetic operators.

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::ldexp(num, exp) has the same effect as std::ldexp(static_cast<double>(num), exp).

For exponentiation of 2 by a floating point exponent, std::exp2 can be used.

#include <cerrno>
#include <cfenv>
#include <cmath>
#include <cstring>
#include <iostream>
// #pragma STDC FENV_ACCESS ON
 
int main()
{
    std::cout
        << "ldexp(5, 3) = 5 * 8 = " << std::ldexp(5, 3) << '\n'
        << "ldexp(7, -4) = 7 / 16 = " << std::ldexp(7, -4) << '\n'
        << "ldexp(1, -1074) = " << std::ldexp(1, -1074)
        << " (minimum positive subnormal float64_t)\n"
        << "ldexp(nextafter(1,0), 1024) = "
        << std::ldexp(std::nextafter(1,0), 1024)
        << " (largest finite float64_t)\n";
 
    // special values
    std::cout << "ldexp(-0, 10) = " << std::ldexp(-0.0, 10) << '\n'
              << "ldexp(-Inf, -1) = " << std::ldexp(-INFINITY, -1) << '\n';
 
    // error handling
    std::feclearexcept(FE_ALL_EXCEPT);
    errno = 0;
    const double inf = std::ldexp(1, 1024);
    const bool is_range_error = errno == ERANGE;
 
    std::cout << "ldexp(1, 1024) = " << inf << '\n';
    if (is_range_error)
        std::cout << "    errno == ERANGE: " << std::strerror(ERANGE) << '\n';
    if (std::fetestexcept(FE_OVERFLOW))
        std::cout << "    FE_OVERFLOW raised\n";
}

Possible output:

ldexp(5, 3) = 5 * 8 = 40
ldexp(7, -4) = 7 / 16 = 0.4375
ldexp(1, -1074) = 4.94066e-324 (minimum positive subnormal float64_t)
ldexp(nextafter(1,0), 1024) = 1.79769e+308 (largest finite float64_t)
ldexp(-0, 10) = -0
ldexp(-Inf, -1) = -inf
ldexp(1, 1024) = inf
    errno == ERANGE: Numerical result out of range
    FE_OVERFLOW raised

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