ldexp, ldexpf, ldexpl - cppreference.com (original) (raw)

Defined in header <math.h>
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)
Defined in header <tgmath.h>
#define ldexp( arg, exp ) (4) (since C99)

1-3) Multiplies a floating-point value arg by the number 2 raised to the exp power.

  1. 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.

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[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