RE: [sv-bc] tf_port_item footnote 33

One interpretation of these two requirements would be as follows:

 

If a prototype contains a named port declaration, the out-of-block
method

declaration must contain an identical port declaration.  

If a prototype contains an anonymous port declaration, then the
out-of-block

method must contain a named port declaration, and that name shall be the

authoritative name for the port (used in named argument association,
etc.).

 

I think this is pretty reasonable, and practical for the user.

 

Regards,

Doug

 

 

From: owner-sv-bc@server.eda.org [mailto:owner-sv-bc@server.eda.org] On
Behalf Of Francoise Martinolle
Sent: Tuesday, July 24, 2007 10:22 AM
To: sv-ec@server.eda-stds.org; sv-bc@server.eda.org
Subject: [sv-bc] tf_port_item footnote 33

 

Someone reported this inconsistency to me:

 

In the SV BNF, there is a note in the tf_port_item production as
follows:

"tf_port_item33 ::=

{ attribute_instance }

[ tf_port_direction ] [ var ] data_type_or_implicit [ port_identifier {
variable_dimension } [ = expression ] ]

33) In a tf_port_item, it shall be illegal to omit the explicit
port_identifier except within a function_prototype or task_prototype."

tf_port_item is used in tf_port_list which is used in task_prototype,
function_prototype and method prototype.

But this seems contradictory with the text for out of block declarations
in 8.22

"The out-of-block method declaration must match the prototype
declaration exactly; the only syntactical difference is that the method
name is preceded by the class name and the class scope resolution
operator ::"

So is the following code illegal?

class foo;

extern function bar ( int);   // prototype

endclass

function foo::bar ( int i ); ... endfunction  // function declaration

If the explicit port_identifier is ONLY optional in a prototype, how can
its implementation (task declaration or function declaration ) match
exactly?

If we believe the BNF and its note, the above extern'd prototype is
legal but its implementation is not.

 

Francoise

    


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