Operator overloading -- lets collect some use cases
smjg_1998 at yahoo.com
Wed Dec 31 05:03:07 PST 2008
> Don wrote:
>> There's been some interesting discussion about operator overloading
>> over the past six months, but to take the next step, I think we need
>> to ground it in reality. What are the use cases?
>> I think that D's existing opCmp() takes care of the plethora of
>> trivial cases where <, >= etc are overloaded. It's the cases where the
>> arithmetic and logical operations are overloaded that are particularly
>> interesting to me.
>> The following mathematical cases immediately spring to mind:
>> * complex numbers
>> * quaternions
>> * vectors
>> * matrices
>> * tensors
>> * bigint operations (including bigint, bigfloat,...)
>> I think that all of those are easily defensible.
>> But I know of very few reasonable non-mathematical uses.
>> In C++, I've seen them used for iostreams, regexps, and some stuff
>> that is quite frankly bizarre.
>> So, please post any use cases which you consider convincing.
> Some observations based on the use cases to date:
> a += b is ALWAYS a = a + b (and likewise for all other operations).
> opXXXAssign therefore seems to be a (limited) performance optimisation.
> The compiler should be allowed to synthesize += from +. This would
> almost halve the minimum number of repetitive functions required.
I thought it did. I must've been imagining it. But the really silly
thing is that, for classes, it seems you can't even make a += b behave
as a = a + b, i.e. reassigning the reference a. Consequently, having
op= on an immutable big integer class is out of the question. There was
a proposal that I supported a while back - see
and followups. The same would work for opXXXAssign (albeit not static).
In short, the way it could be made to work is:
- If a.opAddAssign(b) exists and returns something, make a += b
equivalent to a = a.opAddAssign(b)
- If a.opAddAssign(b) exists and returns void, make a += b equivalent to
(this is a bit like another of my too-much-ignored proposals
- Otherwise, make a += b equivalent to a = a + b.
Of course, the expression represented by a would be evaluated only once
in each case.
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