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Defined in header <cmath> | ||
| (1) | ||
float ldexp(float num,int exp); double ldexp(double num,int exp); | (until C++23) | |
constexpr/* floating-point-type */ ldexp(/* floating-point-type */ num,int exp); | (since C++23) | |
float ldexpf(float num,int exp); | (2) | (since C++11) (constexpr since C++23) |
longdouble ldexpl(longdouble num,int exp); | (3) | (since C++11) (constexpr since C++23) |
Additional overloads(since C++11) | ||
Defined in header <cmath> | ||
template<class Integer> double ldexp( Integer num,int exp); | (A) | (since C++11) (constexpr since C++23) |
std::ldexp for all cv-unqualified floating-point types as the type of the parameternum.(since C++23)A) Additional overloads are provided for all integer types, which are treated asdouble. | (since C++11) |
Contents |
| num | - | floating-point or integer value |
| exp | - | integer value |
If no errors occur,num multiplied by 2 to the power ofexp (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 inmath_errhandling.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
On binary systems (whereFLT_RADIX is2),std::ldexp is equivalent tostd::scalbn.
The functionstd::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 argumentnum of integer type,std::ldexp(num, exp) has the same effect asstd::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 valuesstd::cout<<"ldexp(-0, 10) = "<< std::ldexp(-0.0,10)<<'\n'<<"ldexp(-Inf, -1) = "<< std::ldexp(-INFINITY,-1)<<'\n'; // error handlingstd::feclearexcept(FE_ALL_EXCEPT);errno=0;constdouble inf= std::ldexp(1,1024);constbool 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 = 40ldexp(7, -4) = 7 / 16 = 0.4375ldexp(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) = -0ldexp(-Inf, -1) = -infldexp(1, 1024) = inf errno == ERANGE: Numerical result out of range FE_OVERFLOW raised
(C++11)(C++11) | decomposes a number into significand and base-2 exponent (function)[edit] |
(C++11)(C++11)(C++11)(C++11)(C++11)(C++11) | multiplies a number byFLT_RADIX raised to a power (function)[edit] |
(C++11)(C++11)(C++11) | returns2 raised to the given power (\({\small 2^x}\)2x) (function)[edit] |
C documentation forldexp | |
