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[expr.call] Clarify the value of this #5399

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[expr.call] Clarify the value of this #5399

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10 changes: 6 additions & 4 deletions source/expressions.tex
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Expand Up @@ -3197,10 +3197,12 @@
int x = f<int>(); // error: no argument for second function parameter
\end{codeblock}
\end{example}
If the function is an implicit object member
function, the \keyword{this} parameter of the function\iref{expr.prim.this}
is initialized with a pointer to the object of the call, converted
as if by an explicit type conversion\iref{expr.cast}.
If the function is an implicit object member function,
the object for which it is invoked\iref{expr.prim.this}
is the object which would be pointed to by the result
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Why "would"? There is nothing hypothetical here. The member function is invoked for a certain subobject (or possibly the entire object) of the object designated by the object expression.

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Why "would"? There is nothing hypothetical here.

We don't do any code transformations or anything else to make this conversion «really exist».

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We do convert the object expression to the type of the base class, and the result of that transformation is the value of "this" inside the member function, in my view.

of an explicit type conversion\iref{expr.cast}
on a pointer pointing to the result of the \termref{object expression}
to the type of \keyword{this} in the body of the function.
\begin{note}
There is no access or ambiguity checking on this conversion; the access
checking and disambiguation are done as part of the (possibly implicit)
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