Re: Template argument deduction from member functions...
werasm wrote:
Hi all,
I've found this little piece of code that surprised me when it did not
compile:
struct B1{ void x(); };
struct B2{ void y(); };
struct C : B1, B2{ using B1::x; };
template <class T>
T foo( void (T::*mf)() );
int main()
{
typedef void (C::*Tmf)();
Tmf mf = &C::x;
foo( &C::x ).y(); //Fails - B1 deduced
foo( mf ).y(); //OK
return 0;
}
It surprised me that B1 was deduced at T instead of C. Could anyone
perhaps enlighten me from the standard.
It's simple, really. Expression &C::x looks for 'x' in 'C'. It finds
one, 'B1::x', brought to the 'C' scope by the 'using' declaration. The
type of 'C::x' therefor is actually 'void (B1::*)()'. So, there is no
surprise that when you use '&C::x' directly, 'B1' is deduced.
When you declared 'mf', you force its type to be 'void (C::*)()'. There
is no way for the compiler to know that you're going to assign the same
'B1::x' to it (well, it could conclude it from the logic of the code,
but it chooses to ignore it). So, the compiler is forced to use what it
has, the class of 'mf' is 'C', and so it is deduced.
V
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