Re: Default function parameter values

Victor Bazarov wrote:
> ...
> Set to 0 of what type?  '0' has the type 'int'.  It is convertible
> to any pointer type and to any pointer-to-member type, but the
> compiler needs to know what the destination type is.  It cannot
> know from the type from the expression you supplied, that's why it
> complains.
> ...

Strictly speaking, the compiler cannot deduce the type simply because 
default arguments do not participate in template argument deduction at 
all. The fact that integral constant '0' is convertible to many 
different pointer types plays no role here.

-- 
Best regards,
Andrey Tarasevich