Implement Type is Type
#782
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Relevant lines pasted from the grammar:
```c
//G is-as-expression:
//G prefix-expression
//G is-as-expression is-value-constraint
//GTODO type-id is-type-constraint
```
A *prefix-expression* can be an *unqualified-id*, which can also be a
*type-id*. So when the latter is implemented, it would break any code
using `identifier is` to test an expression.
I understand the usual way of disambiguating is to require parens, so I
have done that here to fix 5 `regression_tests`.
JohelEGP
suggested changes
Oct 25, 2023
| auto is() -> std::false_type { return {}; } | ||
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| template <typename X, typename C> | ||
| requires std::same_as<X, C> || std::derived_from<X,C> |
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Sorry, I was wrong.
std::same_as can be true for non-class types,
whereas std::derived_from can only be true for classes.
Co-authored-by: Johel Ernesto Guerrero Peña <[email protected]> Signed-off-by: Nick Treleaven <[email protected]>
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What about the alternative grammar? |
If this implements #358 (comment), |
|
Hi! Sorry it took me so long to get to this one. I'm going to close it as probably-dated now, but if you want to pursue it please reopen and refresh. Thanks again, and sorry again for the lag. |
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This supplants #759. The same rationale applies for requiring
(identifier) isto test an expression, see that PR for details. I added a diagnostic foridentifier is exprwhereexpris not a type-id, becauseType is expris not valid - instead it will suggest using(identifier) is.cpp2util.hchanges are taken from #701 by @filipsajdak. That makesstd::runtime_error is std::exceptiontrue, which seems correct, although I think that is not mentioned in P2392.I had to add a
speculativeparameter toparser::type_id()which istruewhen there might not be a type (but an expression instead). This avoids an error forptr*which is parsed as a postfix-expression from prefix-expression from is-as-expression.