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[expr.pre] p3 "operators that are overloaded" is not a defined term #6190

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xmh0511 opened this issue Mar 15, 2023 · 4 comments
Open

[expr.pre] p3 "operators that are overloaded" is not a defined term #6190

xmh0511 opened this issue Mar 15, 2023 · 4 comments

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@xmh0511
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xmh0511 commented Mar 15, 2023 •
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[expr.pre] p3 says

Subclause [expr.compound] defines the effects of operators when applied to types for which they have not been overloaded.

We just say a function operator implements an operator, as per [over.oper.general] p1. We didn't define "an operator is/is not overloaded". If a class type does not have any user-defined function operator/template, is it? Moreover, [over.match.oper] p1 says

If no operand of an operator in an expression has a type that is a class or an enumeration, the operator is assumed to be a built-in operator and interpreted according to [expr.compound].

So, [expr.pre] p3 can be changed to

Subclause [expr.compound] defines the effects of operators when applied to types that are neither class nor enumeration type.

This is concise and clear.
@frederick-vs-ja
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I think this shorter form is OK:

[...] when no operator function is selected.

@xmh0511
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xmh0511 commented Mar 16, 2023 •
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[...] when no operator function is selected.

But, the built-in operators for built-in types are not considered to be operator functions.

[over.built] p1

The candidate operator functions that represent the built-in operators defined in [expr.compound] are specified in this subclause.
These candidate functions participate in the operator overload resolution process as described in [over.match.oper] and are used for no other purpose.

@frederick-vs-ja
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But, the built-in operators for built-in types are not considered to be operator functions.

Yes, this is exactly the reason why I suggested the shorter form. I think the sentence should just mean that the effects of built-in operators are covered in [expr.compound].

@xmh0511
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xmh0511 commented Mar 16, 2023 •
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But, the built-in operators for built-in types are not considered to be operator functions.

Yes, this is exactly the reason why I suggested the shorter form. I think the sentence should just mean that the effects of built-in operators are covered in [expr.compound].

"when no operator function is selected" could imply this meaning: there is no viable operator function. However, the built-in types do not have operator functions.

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@xmh0511 @frederick-vs-ja