homsan toft wrote:[color=blue]
> I've tried the below code with MSVC and Comeau online compiler.
> Both complain that operator<< for Outer<part<size_t> >::inner is not
> defined. So how do I declare it without doing full specialization?
>
> I've tried a template on any type, like Outer<AnyT>::inner (see
> below) Also tried a template on the templated type: Outer<part<X>[color=green]
> >::inner[/color]
> // (see below) I can't find a typo? What is the rule in operation here?
>
> Thanks,
> homsan
> ---------------- 8< -----------------
>
> #include <iostream>
>
> template<class SizeT>
> struct part {
> typedef SizeT size_type;
> size_type size;
> };
>
> template<class PartT>
> class Outer
> {
> public:
> typedef PartT part_type;
> struct inner {
> part_type p;
> };
> inner get() const { return inner(); }
> };
>
> // WHY is this not found in main()?[/color]
This is "non-deducible context". You're not using 'PartT' here.
Why do you think you need it? Why couldn't you use
template<class Blah> ..., Blah const& it)
???
[color=blue]
> template<class PartT>
> inline std::ostream& operator<<(std::ostream& ostr, typename
> Outer<PartT>::inner const& it) {
> ostr << it.p.size << " by Jove\n";
> return ostr;
> }
>
> // Nor is this found - what gives?[/color]
It's called "non-deducible context". Why do you need to know 'SizeT'
anyway?
You're not using it. Why could't you simply use OuterInner?
[color=blue]
> template<class SizeT>
> inline std::ostream& operator<<(std::ostream& ostr, typename
> Outer<part<SizeT> >::inner const& it) {
> ostr << it.p.size << " by Jove\n";
> return ostr;
> }
>
> // Now this works, but of course I don't want to redefine for every
> specialization... /*
> std::ostream& operator<<(std::ostream& ostr, Outer<part<size_t>[color=green]
> >::inner const& it) {[/color]
> ostr << it.p.size << " by Jove\n";
> return ostr;
> }
> */
>
> int main() {
> Outer<part<size_t> > yippie;
> std::cout << yippie.get();
> return 0;
> }[/color]
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