pyqgis - Let QGIS 3.0 Processing algorithm output a VectorLayer loaded via the delimited text provider

Following up on this question, I need to be able to setup the QgsVectorLayer loaded via the delimitedtext provider as my algorithm output.

Note that I am still working from this template.

For lack of a better idea, I have defined (in initAlgorithm) the output like so:

self.addParameter(
    QgsProcessingParameterVectorDestination(
        self.OUTPUT,
        self.tr('Output'),
        type = QgsProcessing.TypeVectorAnyGeometry
    )
)

And my Algorithm currently looks like this:

def processAlgorithm(self, parameters, context, feedback):

    ## Define some parameters and functions to handle the loaded file.

    # Getting parameters values to local variables
    crs = self.parameterAsCrs(parameters, self.CRS, context)
    inputFile = self.parameterAsFile(parameters, self.FILEIN, context)
    ver = self.parameterAsEnum(parameters, self.VERSION, context)

    dest_id = self.parameterAsOutputLayer(parameters, self.OUPUT, context)


    uri = ('file:///{}?'
              'type=regexp'
              '&delimiter={}'
              '&skipLines={}'
              '&useHeader=No'
              '&trimFields=Yes'
              '&xField={}'
              '&yField={}'
              '&crs={}'

              '&spatialIndex=Yes'
              '&subsetIndex=no'
              '&watchFile=Yes').format(inputFile,
                                       regex,
                                       skips,
                                       'Easting',
                                       'Northing',
                                       crs.geographicCrsAuthId())

    vl = QgsVectorLayer(uri, 'MY_LAYER',  'delimitedtext')


    # Setup Fields Aliases
    for idx, alias in enumerate(myFields):
        try:
            vl.setFieldAlias(idx, alias)
        except:
            continue

    return {'MY_LAYER': dest_id}

The return line is where the crash happens but I'm running out of ideas: by comparison with the template where self.parameterAsSink() returned a tuple (sink, dest_id), I'm guessing the self.parameterAsOutputLayer() returns the dest_id.
Then I try to return the created layer vl and associate it with the dest_id, again, by analogy with the template.

I am stuck here as Qgis (3.0.2) crashes (right at the return line, every line of code gets executed without problem) every time I try to run this current code (but I can make it work fine as a standalone script from the Python Console...)

I have tried the following variations and all lead to invariable crashing:
return 'MY_LAYER'
return {vl: dest_id}
return {dest_id}
return dest_id
return vl
return

Answer

Because your output here is effectively hard-coded, you don't need to define it as a input parameter. Instead declare it as an output that your algorithm generates:

self.addOutput(
    QgsProcessingOutputVectorLayer(
        self.OUTPUT,
        self.tr('Output'),
        type = QgsProcessing.TypeVectorAnyGeometry)
)

Then, in your processAlgorithm method (simplified to just the relevant bits):

def processAlgorithm(self, parameters, context, feedback):
    crs = self.parameterAsCrs(parameters, self.CRS, context)
    inputFile = self.parameterAsFile(parameters, self.FILEIN, context)

    uri = ('file:///{}?'
           'type=regexp'
           '&delimiter={}'
           '&skipLines={}'
           '&useHeader=No'
           '&trimFields=Yes'

           '&xField={}'
           '&yField={}'
           '&crs={}'
           '&spatialIndex=Yes'
           '&subsetIndex=no'
           '&watchFile=Yes').format(inputFile,
                                    regex,
                                    skips,
                                    'Easting',
                                    'Northing',

                                    crs.geographicCrsAuthId())

    vl = QgsVectorLayer(uri, 'MY_LAYER', 'delimitedtext')

    return {self.OUTPUT: vl}