[expr.pre] p3 "operators that are overloaded" is not a defined term
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Description
[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.
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Research direction
Read [expr.pre] p3 alongside [expr.compound], [over.oper.general], and [over.match.oper] to confirm the terminology and scope. Update the sentence in [expr.pre] to use a defined distinction, then check the rendered wording and relevant cross-references in the draft.
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