ldexp, ldexpf, ldexpl

[edit] Parameters

[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),

• 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 arg is ±0, it is returned, unmodified
• If arg is ±∞, it is returned, unmodified
• If exp is 0, then arg is returned, unmodified
• If arg is NaN, NaN is returned.

[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");
}

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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