Please tell me if am right (and that VS6 is wrong). Defining a function's
parameter to be const is completely irrelevant except in the actual
implementation. Thus, the following should work according to the Standard.
Thanks.

void foo( void (*)(int) );
void bar( int const );

void test() {foo( bar );}



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In article <39ef6c88$1_2@news.nwlink.com>,
Gary Hinger <garyh@zipperint.com> wrote:
>Please tell me if am right (and that VS6 is wrong). Defining a function's
>parameter to be const is completely irrelevant except in the actual
>implementation.

Right, at the top level that is.

>Thus, the following should work according to the Standard.
>
>void foo( void (*)(int) );
>void bar( int const );
>
>void test() {foo( bar );}

Yes.  Actually, it doesn't seem to come into play in this example
no matter what.

- Greg
--
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In article <39ef6c88$1_2@news.nwlink.com>, Gary Hinger wrote:
>Please tell me if am right (and that VS6 is wrong). Defining a function'=
s
>parameter to be const is completely irrelevant except in the actual
>implementation. Thus, the following should work according to the Standar=
d.
>Thanks.
>
>void foo( void (*)(int) );
>void bar( int const );
>
>void test() {foo( bar );}
[-]
8.3.5 Functions
---------------

3) ...=20
   Any cv-qualifier modifying a parameter type is deleted. [Example:
   the type void (*)( const int) becomes void (*)(int) -- end example]

   Such cv-qualifiers affect only the definition of the parameter within
   the the body of the function; they do not affect the function type.
   ...

Cheers,
Juergen

--=20
\ Real name     : J=FCrgen Heinzl         \       no flames      /
 \ EMail Private : monocerus@n-online.de \ send money instead /

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