 `x` can be destroyed; the default is False.
        See `fft` for more details.
    workers : int, optional
        Maximum number of workers to use for parallel computation. If negative,
        the value wraps around from ``os.cpu_count()``.
        See :func:`~scipy.fft.fft` for more details.
    plan : object, optional
        This argument is reserved for passing in a precomputed plan provided
        by downstream FFT vendors. It is currently not used in SciPy.

        .. versionadded:: 1.5.0

    Returns
    -------
    out : ndarray
        The truncated or zero-padded input, transformed along the axis
        indicated by `axis`, or the last one if `axis` is not specified.
        The length of the transformed axis is `n`, or, if `n` is not given,
        ``2*m - 2``, where ``m`` is the length of the transformed axis of
        the input. To get an odd number of output points, `n` must be
        specified, for instance, as ``2*m - 1`` in the typical case,

    Raises
    ------
    IndexError
        If `axis` is larger than the last axis of `a`.

    See Also
    --------
    rfft : Compute the 1-D FFT for real input.
    ihfft : The inverse of `hfft`.
    hfftn : Compute the N-D FFT of a Hermitian signal.

    Notes
    -----
    `hfft`/`ihfft` are a pair analogous to `rfft`/`irfft`, but for the
    opposite case: here the signal has Hermitian symmetry in the time
    domain and is real in the frequency domain. So, here, it's `hfft`, for
    which you must supply the length of the result if it is to be odd.
    * even: ``ihfft(hfft(a, 2*len(a) - 2) == a``, within roundoff error,
    * odd: ``ihfft(hfft(a, 2*len(a) - 1) == a``, within roundoff error.

    Examples
    --------
    >>> from scipy.fft import fft, hfft
    >>> import numpy as np
    >>> a = 2 * np.pi * np.arange(10) / 10
    >>> signal = np.cos(a) + 3j * np.sin(3 * a)
    >>> fft(signal).round(10)
    array([ -0.+0.j,   5.+0.j,  -0.+0.j,  15.-0.j,   0.+0.j,   0.+0.j,
            -0.+0.j, -15.-0.j,   0.+0.j,   5.+0.j])
    >>> hfft(signal[:6]).round(10) # Input first half of signal
    array([  0.,   5.,   0.,  15.,  -0.,   0.,   0., -15.,  -0.,   5.])
    >>> hfft(signal, 10)  # Input entire signal and truncate
    array([  0.,   5.,   0.,  15.,  -0.,   0.,   0., -15.,  -0.,   5.])
    r