Re: [Boost-users] problem with boost::is_same

Subject: Re: [Boost-users] problem with boost::is_same
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2009-01-17 22:13:37

AMDG

Zachary Turner wrote:
> I have a question about the following code:
>
>     typedef void(intArrayParamFunc)(int[]);
>     typedef void(intPtrParamFunc)(int*);
>
>     
> BOOST_STATIC_ASSERT((boost::is_same<intArrayParamFunc,intPtrParamFunc>::value));
>     BOOST_STATIC_ASSERT((boost::is_same<int*,int[]>::value));
>
>
>     cout << typeid(intArrayParamFunc).name() << endl;
>     cout << typeid(intPtrParamFunc).name() << endl;
>
>     cout << typeid(int*).name() << endl;
>     cout << typeid(int[]).name() << endl;
>
> The first assert compiles but the second fails, which means that 
> boost::is_same<> thinks that int* and int[] are different, but 
> void(int[]) and void(int*) are the same.  Why is this?
>
> The typeids print out the following:
>
> void __cdecl(int * const)
> void __cdecl(int *)
> int *
> int [0]
>
> I'm using Visual Studio 2008.  I notice that typeid tells me that 
> intArrayParamFunc is void(int* const).  So why does boost::is_same<> 
> compare them equal?  Surely void(int* const) should not be the same 
> type as void(int*).

Yes, it is the same.  They are the same because the two
signatures are equivalent at the call site.  The only
difference between

void f(const int);
and
void f(int);

is that in the first case f can modify its arguments.  Since
the argument is passed by value, this has no impact on the caller.

> Unless this is just some obscure language subtlety I'm not aware of.  
> So is this a bug in boost::is_same<>, a bug in my compiler, or no bugs 
> at all and everything is working perfectly?

Everything is working perfectly.  An array argument to a function is
really a pointer.

In Christ,
Steven Watanabe