def fadd(ctx, x, y, **kwargs):
        """
        Adds the numbers *x* and *y*, giving a floating-point result,
        optionally using a custom precision and rounding mode.

        The default precision is the working precision of the context.
        You can specify a custom precision in bits by passing the *prec* keyword
        argument, or by providing an equivalent decimal precision with the *dps*
        keyword argument. If the precision is set to ``+inf``, or if the flag
        *exact=True* is passed, an exact addition with no rounding is performed.

        When the precision is finite, the optional *rounding* keyword argument
        specifies the direction of rounding. Valid options are ``'n'`` for
        nearest (default), ``'f'`` for floor, ``'c'`` for ceiling, ``'d'``
        for down, ``'u'`` for up.

        **Examples**

        Using :func:`~mpmath.fadd` with precision and rounding control::

            >>> from mpmath import *
            >>> mp.dps = 15; mp.pretty = False
            >>> fadd(2, 1e-20)
            mpf('2.0')
            >>> fadd(2, 1e-20, rounding='u')
            mpf('2.0000000000000004')
            >>> nprint(fadd(2, 1e-20, prec=100), 25)
            2.00000000000000000001
            >>> nprint(fadd(2, 1e-20, dps=15), 25)
            2.0
            >>> nprint(fadd(2, 1e-20, dps=25), 25)
            2.00000000000000000001
            >>> nprint(fadd(2, 1e-20, exact=True), 25)
            2.00000000000000000001

        Exact addition avoids cancellation errors, enforcing familiar laws
        of numbers such as `x+y-x = y`, which don't hold in floating-point
        arithmetic with finite precision::

            >>> x, y = mpf(2), mpf('1e-1000')
            >>> print(x + y - x)
            0.0
            >>> print(fadd(x, y, prec=inf) - x)
            1.0e-1000
            >>> print(fadd(x, y, exact=True) - x)
            1.0e-1000

        Exact addition can be inefficient and may be impossible to perform
        with large magnitude differences::

            >>> fadd(1, '1e-100000000000000000000', prec=inf)
            Traceback (most recent call last):
              ...
            OverflowError: the exact result does not fit in memory

        """
        prec, rounding = ctx._parse_prec(kwargs)
        x = ctx.convert(x)
        y = ctx.convert(y)
        try:
            if hasattr(x, '_mpf_'):
                if hasattr(y, '_mpf_'):
                    return ctx.make_mpf(mpf_add(x._mpf_, y._mpf_, prec, rounding))
                if hasattr(y, '_mpc_'):
                    return ctx.make_mpc(mpc_add_mpf(y._mpc_, x._mpf_, prec, rounding))
            if hasattr(x, '_mpc_'):
                if hasattr(y, '_mpf_'):
                    return ctx.make_mpc(mpc_add_mpf(x._mpc_, y._mpf_, prec, rounding))
                if hasattr(y, '_mpc_'):
                    return ctx.make_mpc(mpc_add(x._mpc_, y._mpc_, prec, rounding))
        except (__HOLE__, OverflowError):
            raise OverflowError(ctx._exact_overflow_msg)
        raise ValueError("Arguments need to be mpf or mpc compatible numbers")
dataset/ETHPy150Open fredrik-johansson/mpmath/mpmath/ctx_mp.py/MPContext.fadd