On Fri, Jul 29, 2022 at 8:15 PM Jason McKesson via Std-Discussion <std-discussion@lists.isocpp.org> wrote:
This question is purely theoretical. Let's say we for some reason have
this type:

struct conv
  template<typename T>
  operator some_type();

It has a conversion operator which is a template. However, the type
the conversion operator converts to is not related to the template
parameter of the function. So if we want to call that function, we
would have to do so explicitly by providing a template parameter.

That would look like this:

test t;
auto temp = t.operator some_type<other_type>();

One problem though. [class.conv.fct]/3 says:

> The conversion-type-id in a conversion-function-id is the longest sequence of tokens that could possibly form a conversion-type-id.

The token sequence "some_type<other_type>" "could possibly form a
conversion-type-id". And therefore, it is taken as an entire type ID.
This means that the template arguments are taken as part of the type
ID, and *not* as arguments to the conversion operator template

I think this is incorrect; some_type<other_type> cannot form a conversion-type-id because some_type<other_type> is not a valid type-specifier. A simple-template-id is a type-specifier, but takes the form template-name<template-argument-list(opt)>. Here, some_type is not a template-name, because there isn't any template named some_type.

However, it appears that you're correct that there is no syntax that allows specifying explicit template arguments to a conversion function. The syntax of template-id does not permit it.

So... is there a way to work around this? Can you actually call such a
function, or is this a corner case where C++ allows you to write a
function which cannot be called?

I would argue that the work-around is to simply not declare such a thing. You obviously aren't intending to use it the way that a conversion function is typically used (i.e., for an implicit or explicit conversion), so you don't need it to be a conversion function in the first place.

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Brian Bi