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      std::modf,std::modff,std::modfl

      From cppreference.com
      <cpp‎ |numeric‎ |math
       
       
       
      Common mathematical functions
      Nearest integer floating point operations
      (C++11)(C++11)(C++11)
      (C++11)
      (C++11)
      (C++11)(C++11)(C++11)
      Floating point manipulation functions
      (C++11)(C++11)
      (C++11)
      (C++11)
      modf
      (C++11)(C++11)
      (C++11)
      Classification and comparison
      (C++11)
      (C++11)
      (C++11)
      (C++11)
      (C++11)
      (C++11)
      Types
      (C++11)
      (C++11)
      (C++11)
      Macro constants
       
      Defined in header<cmath>
      (1)
      float       modf(float num,float* iptr);

      double      modf(double num,double* iptr);

      longdouble modf(longdouble num,longdouble* iptr);
      (until C++23)
      constexpr/* floating-point-type */

                  modf(/* floating-point-type */ num,

                         /* floating-point-type */* iptr);
      (since C++23)
      float       modff(float num,float* iptr);
      (2)(since C++11)
      (constexpr since C++23)
      longdouble modfl(longdouble num,longdouble* iptr);
      (3)(since C++11)
      (constexpr since C++23)
      Defined in header<cmath>
      template<class Integer>
      double      modf( Integer num,double* iptr);
      (A)(constexpr since C++23)
      1-3) Decomposes given floating point valuenum into integral and fractional parts, each having the same type and sign asnum. The integral part (in floating-point format) is stored in the object pointed to byiptr. The library provides overloads ofstd::modf for all cv-unqualified floating-point types as the type of the parameternum and the pointed-to type ofiptr.(since C++23)
      A) Additional overloads are provided for all integer types, which are treated asdouble.
      (since C++11)

      Contents

      [edit]Parameters

      num - floating-point or integer value
      iptr - pointer to floating-point value to store the integral part to

      [edit]Return value

      If no errors occur, returns the fractional part ofnum with the same sign asnum. The integral part is put into the value pointed to byiptr.

      The sum of the returned value and the value stored in*iptr givesnum (allowing for rounding).

      [edit]Error handling

      This function is not subject to any errors specified inmath_errhandling.

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

      • Ifnum is ±0, ±0 is returned, and ±0 is stored in*iptr.
      • Ifnum is ±∞, ±0 is returned, and ±∞ is stored in*iptr.
      • Ifnum is NaN, NaN is returned, and NaN is stored in*iptr.
      • The returned value is exact,the current rounding mode is ignored.

      [edit]Notes

      This function behaves as if implemented as follows:

      double modf(double num,double* iptr){#pragma STDC FENV_ACCESS ONint save_round=std::fegetround();std::fesetround(FE_TOWARDZERO);*iptr=std::nearbyint(num);std::fesetround(save_round);returnstd::copysign(std::isinf(num)?0.0: num-(*iptr), num);}

      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::modf(num, iptr) has the same effect asstd::modf(static_cast<double>(num), iptr).

      [edit]Example

      Compares different floating-point decomposition functions:

      Run this code
      #include <cmath>#include <iostream>#include <limits> int main(){double f=123.45;std::cout<<"Given the number "<< f<<" or "<<std::hexfloat<< f<<std::defaultfloat<<" in hex,\n"; double f3;double f2= std::modf(f,&f3);std::cout<<"modf() makes "<< f3<<" + "<< f2<<'\n'; int i;    f2=std::frexp(f,&i);std::cout<<"frexp() makes "<< f2<<" * 2^"<< i<<'\n';     i=std::ilogb(f);std::cout<<"logb()/ilogb() make "<< f/std::scalbn(1.0, i)<<" * "<<std::numeric_limits<double>::radix<<"^"<<std::ilogb(f)<<'\n'; // special values    f2= std::modf(-0.0,&f3);std::cout<<"modf(-0) makes "<< f3<<" + "<< f2<<'\n';    f2= std::modf(-INFINITY,&f3);std::cout<<"modf(-Inf) makes "<< f3<<" + "<< f2<<'\n';}

      Possible output:

      Given the number 123.45 or 0x1.edccccccccccdp+6 in hex,modf() makes 123 + 0.45frexp() makes 0.964453 * 2^7logb()/ilogb() make 1.92891 * 2^6modf(-0) makes -0 + -0modf(-Inf) makes -INF + -0

      [edit]See also

      (C++11)(C++11)(C++11)
      nearest integer not greater in magnitude than the given value
      (function)[edit]
      Retrieved from "https://en.cppreference.com/mwiki/index.php?title=cpp/numeric/math/modf&oldid=177737"

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