TITLE: cv qualifiers and function overloading

(Newsgroups: comp.lang.c++.moderated, 11 Jun 98)


FVALI: fvali  <fvali@biotrack.com>

>> > void f(int);
>> >
>> > void f(const int);


OLIVA: Alexandre Oliva <oliva@dcc.unicamp.br>

>> These are two equivalent declarations of the same function, with
>> signature void(int), not an example of overloading.  The outermost
>> cv-qualification affects the argument inside the function body, but
>> not the signature of the function [dcl.fct]/3.


FVALI: fvali@my-dejanews.com

>In article <6lgmjg$jmj@netlab.cs.rpi.edu>,
>Yet don't f(int&) and f(const int&) produce different signatures?


CLAMAGE: clamage@Eng.Sun.COM (Steve Clamage)

Yes, but for different reasons.


FVALI:

>The Dec 96 draft says this:
>[Begin Draft Quote:
>8.3.5(3)
>The type of a function is determined using the follow-
>  ing  rules.   The  type  of  each parameter is determined from its own
>  decl-specifier-seq and declarator.  After determining the type of each
>  parameter,  any  parameter of type "array of T" or "function returning
>  T" is adjusted to be "pointer to T" or "pointer to function  returning
>  T,"  respectively.   After producing the list of parameter types, sev-
>  eral transformations take place upon  these  types  to  determine  the
>  function  type.   Any  cv-qualifier  modifying  a  parameter  type  is
>  deleted; e.g., the type void(const int) becomes void(int).   Such  cv-
>  qualifiers only affect the definition of the parameter within the body
>  of the function; they do not affect the function type.
>:End Draft Quote]


CLAMAGE:

The cv-qualifiers that are removed are the "outermost", i.e. the
ones that apply to the parameter itself. In "const int", the
const modifies the type of the parameter. In "const int&", the
const modifies the type that the parameter refers to.

Probably the wording in the draft could be better.