std::modf,s...f,std::modfl(3)

🌸 NAME 🌸

okay so bestie, welcome to the std::modf, std::modff, std::modfl page from the C++ Standard Library manual — and it is literally the glow-up of splitting a floating-point number into its integral and fractional parts 💅✨

The official NAME line is: std::modf,std::modff,std::modfl - std::modf,std::modff,std::modfl.

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🦋 Synopsis 🦋

Defined in header <cmath> — yes, that one, bestie 🎀

          Defined in header <cmath>
          float       modf ( float num, float* iptr );

          double      modf ( double num, double* iptr );               (until
       C++23)

          long double modf ( long double num, long double*
          iptr );
          constexpr /* floating-point-type */

                      modf ( /* floating-point-type */ num,            (since
       C++23)
                                                               (1)
                             /* floating-point-type */* iptr
          );
          float       modff( float num, float* iptr );             (2) (since
       C++11)
                                                                       (constexpr
       since C++23)
          long double modfl( long double num, long double*         (3) (since
       C++11)
          iptr );
       (constexpr since C++23)
          Additional overloads (since C++11)
          Defined in header <cmath>
          template< class Integer >                                (A)
       (constexpr since C++23)
          double      modf ( Integer num, double* iptr );

So 1-3) decompose the given floating-point value num into integral and fractional parts, each having the same type and sign as num 🫶 The integral part (in floating-point format) is stored in the object pointed to by iptr.

The library provides overloads of std::modf for all cv-unqualified floating-point types as the type of the parameter num and the pointed-to type of iptr. (since C++23) ✨

And A) — additional overloads are provided for all integer types, which are treated as double (since C++11) 💖

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🎀 Parameters 🎀

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

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🌈 Return value 🌈

If no errors occur, returns the fractional part of num with the same sign as num ⭐ The integral part is put into the value pointed to by iptr.

The sum of the returned value and the value stored in *iptr gives num (allowing for rounding). no cap, that's the whole vibe 💫

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🔥 Error handling 🔥

This function is not subject to any errors specified in math_errhandling.

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

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

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🧚 Notes 🧚

This function behaves as if implemented as follows — copy-paste energy, but make it exact, bestie:

double modf(double num, double* iptr)
{
#pragma STDC FENV_ACCESS ON
    int save_round = std::fegetround();
    std::fesetround(FE_TOWARDZERO);
    *iptr = std::nearbyint(num);
    std::fesetround(save_round);
    return std::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 argument num of integer type, std::modf(num, iptr) has the same effect as std::modf(static_cast<double>(num), iptr). periodt 💅

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🍓 Example 🍓

Compares different floating-point decomposition functions:

#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 + 0.45
frexp() makes 0.964453 * 2^7
logb()/ilogb() make 1.92891 * 2^6
modf(-0) makes -0 + -0
modf(-Inf) makes -INF + -0

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👑 See also 👑

The see-also shoutouts are exactly:

          trunc
          truncf
          truncl  nearest integer not greater in magnitude than the given value
          (C++11) (function)
          (C++11)
          (C++11)
          C documentation for
          modf

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xoxo, sourced from http://cppreference.com, version/date 2024.06.10, page std::modf,s...f,std::modfl(3) 💖🦋✨