[sv-bc] RE: [sv-ec] overloading an assignment operator

Oops, you are right.  Assignment *is* listed in the overload_operator
production.  It looked to me like a ==, since it was after the other
comparison operators.  Sorry about that.

But this means that there is a contradiction in the LRM.  It explicitly
says that it does not change the meaning of the operators for those data
types where it is legal to apply them.  So you could claim that it works
for assignments where there is no legal implicit cast between the two
types, and provides an explicit cast.  But the example you give includes
assignments between integer and floating point types.  Those already 
have implicit casts defined, and are perfectly legal.

There are also inherent problems with any kind of overloading that
depends on the exact type of any integral expression.  Verilog does so
much implicit conversion in integral expressions that I don't see any
way to determine the exact type of any integral expression.  This is
presumably why overloading is only allowed for other types, such as
unpacked structs.

So I guess we must conclude that overloading of assignment is legal,
but only to provide a cast between two types that are not assignment
compatible (and while it doesn't say so, neither type should be an
integral type, due to the impossibility of determining such types
in general, unless all integral types are considered equivalent for
this purpose).  And then we must conclude that the examples of
assignment overloading are wrong, since the types involved are 
assignment compatible.

So getting back to your original question...

Someone else might be able to respond more authoritatively, but here
is how it looks to me.

No, I don't think assignment would be a binary operator.  It is
essentially providing a cast, not the assignment itself.  You
would not have a function with an input and output argument, that
assigns the input to the output.  Instead, you have a function that
takes the RHS as an input argument, and returns a value of the LHS
type, which is then assigned to the LHS.  So it is a unary operator.

The examples seem to support this, despite their other problems.

Steven Sharp
sharp@cadence.com


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