[cxx-abi-dev] mangling for fold-expressions

[John McCall]

On Nov 10, 2014, at 11:17 AM, Richard Smith <richardsmith at google.com> wrote:
> On 10 November 2014 11:13, John McCall <rjmccall at apple.com> wrote:
> On Nov 10, 2014, at 10:56 AM, Richard Smith <richardsmith at google.com> wrote:
>> On 10 November 2014 10:30, John McCall <rjmccall at apple.com> wrote:
>> On Nov 8, 2014, at 9:16 AM, Richard Smith <richardsmith at google.com> wrote:
>> > Hi,
>> >
>> > WG21 is voting on a proposal for "fold-expressions" today. These are syntactically of the form:
>> >
>> >   ( .... + pack )
>> >   ( pack + ... )
>> >   ( p0 + ... + pack )
>> >   ( pack + ... + pn )
>> >
>> > (where + can be any binary operator). These expand to
>> >
>> >   (((p0 + p1) + ...) + pn)
>> >
>> > for the first and third cases and
>> >
>> >   (p0 + (p1 + (... + pn)))
>> >
>> > for the other two cases.
>> 
>> I should add:
>>  - in the first and second cases, the pack is p0 ... pn
>>  - in the third case, the pack is p1 ... pn
>>  - in the fourth case, the pack is p0 ... p{n-1}
> 
> Your notation is terrible, Richard. :)
> 
> Yeah, sorry. I should have just attached the paper rather than trying to summarize. (Now attached.)
>  
> Okay, so the idea is that:
>   E op … op P => (((E op P1) op P2) op …) op Pn
>   P op … op E => P1 op (P2 op (… op (Pn op E)))
> 
> How are E and P determined?  Everything preceding/following the … term, or does this actually follow the grammar’s associativity rules if you had e.g.
>   1+2+…+packref+3+4
> or does it only work within parentheses?
> 
> Parentheses are required, and only cast-expressions are allowed before / after the operators, so there are no precedence / associativity issues. (The initial proposal was deliberately very conservative in this regard.)

Okay, makes sense.

>> > These need a mangling; I suggest (and have implemented):
>> >
>> > <expression> ::=
>> >   fl <binary operator-name> <expression>   # ( ... op pack )
>> >   fr <binary operator-name> <expression>   # ( pack op ... )
>> >   fx <binary operator-name> <expression> <expression>   # ( expr op ... op expr )
>> 
>> This doesn’t seem to correspond to one of your examples.
>> 
>> You get this for the third and fourth cases.
> 
> Okay, and you’re saying that left/right folding are disambiguated because one of the expressions contains a pack and the other doesn’t?  I think I would prefer this to be explicit in the mangling.
> 
> OK. Then fl / fr / fL / fR?

Sounds good to me.

John.
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