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# std::modf

From cppreference.com
< cpp‎ | numeric‎ | math

Defined in header `<cmath>` |  |
float modf( float x, float* iptr ); | (1) |
double modf( double x, double* iptr ); | (2) |
long double modf( long double x, long double* iptr ); | (3) |

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

### [edit] Parameters

x | - | floating point value
iptr | - | pointer to floating point value to store the integral part to

### [edit] Return value

If no errors occur, returns the fractional part of `x` with the same sign as `x`. 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 `x` (allowing for rounding)

### [edit] Error handling

This function is not subject to any errors specified in math_errhandling

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

- If `x` is ±0, ±0 is returned, and ±0 is stored in *iptr.
- If `x` is ±∞, ±0 is returned, and ±∞ is stored in *iptr.
- If `x` 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 x, double* iptr)
    {
    #pragma STDC FENV_ACCESS ON
        int save_round = std::fegetround();
        std::fesetround(FE_TOWARDZERO);
        *iptr = std::nearbyint(x);
        std::fesetround(save_round);
        return std::copysign(std::isinf(x) ? 0.0 : x - (*iptr), x);
    }

### [edit] Example

Compares different floating-point decomposition functions

Run this code

    #include <iostream>
    #include <cmath>
    #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

### [edit] See also

trunc (C++11) | nearest integer not greater in magnitude than the given value (function) [edit]
C documentation for modf

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