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

ldexp, ldexpf, ldexpl - cppreference.com

float       ldexpf( float arg, int exp ); (1) (since C99) double      ldexp( double arg, int exp ); (2) long double ldexpl( long double arg, int exp ); (3) (since C99)

#define ldexp( arg, exp )

(4) (since C99) 1-3)

Multiplies a floating-point value

arg

by the number

2

raised to the

exp

power.

4) Type-generic macro: If arg has type long double, ldexpl is called. Otherwise, if arg has integer type or the type double, ldexp is called. Otherwise, ldexpf is called, respectively.

[edit] Parameters arg - floating-point value exp - integer value [edit] Return value

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

[edit] Error handling

Errors are reported as specified in math_errhandling.

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

[edit] Notes

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

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

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

[edit] Example
#include <errno.h>
#include <fenv.h>
#include <float.h>
#include <math.h>
#include <stdio.h>
// #pragma STDC FENV_ACCESS ON
 
int main(void)
{
    printf("ldexp(7, -4) = %f\n", ldexp(7, -4));
    printf("ldexp(1, -1074) = %g (minimum positive subnormal double)\n",
            ldexp(1, -1074));
    printf("ldexp(nextafter(1,0), 1024) = %g (largest finite double)\n",
            ldexp(nextafter(1,0), 1024));
 
    // special values
    printf("ldexp(-0, 10) = %f\n", ldexp(-0.0, 10));
    printf("ldexp(-Inf, -1) = %f\n", ldexp(-INFINITY, -1));
 
    // error handling
    errno = 0; feclearexcept(FE_ALL_EXCEPT);
    printf("ldexp(1, 1024) = %f\n", ldexp(1, 1024));
    if (errno == ERANGE)
        perror("    errno == ERANGE");
    if (fetestexcept(FE_OVERFLOW))
        puts("    FE_OVERFLOW raised");
}

Possible output:

ldexp(7, -4) = 0.437500
ldexp(1, -1074) = 4.94066e-324 (minimum positive subnormal double)
ldexp(nextafter(1,0), 1024) = 1.79769e+308 (largest finite double)
ldexp(-0, 10) = -0.000000
ldexp(-Inf, -1) = -inf
ldexp(1, 1024) = inf
    errno == ERANGE: Numerical result out of range
    FE_OVERFLOW raised
[edit] References
[edit] See also

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