<table border="1" cellspacing="0" cellpadding="8">
    <tr>
        <th>Issue</th>
        <td>
            <a href=https://github.com/llvm/llvm-project/issues/76649>76649</a>
        </td>
    </tr>

    <tr>
        <th>Summary</th>
        <td>
            [Clang] operator== rewrite return type restriction is overly strict
        </td>
    </tr>

    <tr>
      <th>Labels</th>
      <td>
            clang
      </td>
    </tr>

    <tr>
      <th>Assignees</th>
      <td>
      </td>
    </tr>

    <tr>
      <th>Reporter</th>
      <td>
          ednolan
      </td>
    </tr>
</table>

<pre>
    [over.match.oper](https://eel.is/c++draft/over.match.oper#10) states: "If a rewritten operator== candidate is selected by overload resolution for an operator @, its return type shall be cv bool."

Presumably based on that language, Clang rejects the following code:

```
struct foo {};

template <typename T>
auto operator==(foo, T) { return true; }

void bar() {
    0 == foo{};
}
```

With this error:

```
<source>:7:7: error: return type 'auto' of selected 'operator==' function for rewritten '==' comparison is not 'bool'
    7 |     0 == foo{};
      |     ~ ^ ~~~~~
<source>:4:6: note: declared here
    4 | auto operator==(foo, T) { return true; }
      |      ^
```

However, GCC, MSVC and EDG all accept it, as shown in Compiler Explorer here: https://godbolt.org/z/46aaPvn46

I think Clang is being overly restrictive here. The fact that the `operator==` returns `auto` doesn't violate that "its return type shall be cv `bool`," since `auto` is a placeholder for a return type, not a return type itself. The standard seems to make this distinction in [dcl.spec.auto](https://eel.is/c++draft/dcl.spec.auto#general-3):

> A placeholder type can appear with a function declarator in the decl-specifier-seq, type-specifier-seq, conversion-function-id, or trailing-return-type, in any context where such a declarator is valid. If the function declarator includes a trailing-return-type ([dcl.fct]), that trailing-return-type specifies the declared return type of the function. Otherwise, the function declarator shall declare a function. If the declared return type of the function contains a placeholder type, the return type of the function is deduced from non-discarded return statements, if any, in the body of the function ([stmt.if]).

Notice the distinction between "declared return type" (which is `auto` or `decltype(auto)`) and "return type" (which is "deduced from non-discarded return statements [...] in the body of the function").

</pre>
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