std:: ldexp, std:: ldexpf, std:: ldexpl

Parameters

Return value

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

Error handling

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.

Notes

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.

Example

#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";
}

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

See also