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Showing content from https://en.cppreference.com/w/cpp/language/../error/error_code/../../../cpp/numeric/math/modf.html below:

std::modf, std::modff, std::modfl - cppreference.com

float       modff( float num, float* iptr );

long double modfl( long double num, long double* iptr );

template< class Integer >
double      modf ( Integer num, double* iptr );

1-3) Decomposes 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)

A) Additional overloads are provided for all integer types, which are treated as double.

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).

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

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

• 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.

This function behaves as if implemented as follows:

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).

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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