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Subject: Re: [boost] [type_traits][function_types] Discard param const qualification, bug or feature?
From: Jonathan Wakely (jwakely.boost_at_[hidden])
Date: 2013-10-01 03:30:02

On 1 October 2013 05:45, Mostafa wrote:
> On Mon, 30 Sep 2013 19:54:56 -0700, Gavin Lambert wrote:
>> A simple add_reference is obviously the wrong thing to be doing then.


>> In general, you can take any parameter type T and wrap it as a "const T&"
>> (or if you prefer, "T const&") and it will do the right thing, unless T was
>> already a reference. But note that you must use a const reference -- a
>> non-const reference won't work.
> Not in my particular use case. The code I was working is generated by a mix
> of PPMP and TMP techniques. The argument x is eventually forwarded to some
> user block of code, where the user expects it to be of the same type as the
> one he/she specified in the signature of SomeUserClass::foo. So had the user
> specified this instead:
> struct SomeUserClass
> {
> static void foo(int x)
> {
> SomeCodeGenClass::foo(x);
> }
> };
> Then it is expected that x would be mutable in that forwarded-to block of
> code.

Well then you're doing it wrong. You should base the forwarded
parameter type on decltype(x) not the parameter type in the function
signature. The function signature doesn't include the top-level
const, that's how C++ works.

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