Template typedef of a template

Question

Possible Duplicate:
C++ template typedef

I am trying to derive a template type of another template by pre-specializing of another template:

template<unsigned a, unsigned b, unsigned c>
struct test
{
    enum
    {
        TEST_X = a,
        TEST_Y = b,
        TEST_Z = c,
    };
};

template<unsigned c>
typedef test<0, 1, c> test01;

However, on GCC 4.4.5, I am getting this error: error: template declaration of ‘typedef’ on the second type (test01).

Guidance would be highly appreciated, as I don't understand what is wrong with my code.

I know this is a duplicate, but I can't find it. EDIT: Aha, [here](http://stackoverflow.com/questions/2795023/c-template-typedef). — GManNickG, Feb 15 '11 at 08:23
When you look at the duplicate, note that in that particular case the user did not want to use inheritance, even if that was a common pattern for C++98 compilers: `template struct test01 : test<0,1,c> {}` There is a slight difference, as you are adding inheritance to the mix and that has some pitfalls (beware: do not delete the type through pointers to `test<0,1,N>` as that will cause UB, always destroy at the `test01` level) — David Rodríguez - dribeas, Feb 15 '11 at 08:59

score 10 · Accepted Answer · answered Feb 15 '11 at 08:23

10

This syntax isn't allowed by C++03. The nearest work-around is:

template<unsigned c>
struct test01
{
    typedef test<0, 1, c> type;
};

typedef test01<2>::type my_type;

In C++0x, we can do this:

template<unsigned c>
using test01 = test<0, 1, c>;

answered Feb 15 '11 at 08:23

GManNickG

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Damn you C++03!!! and damn you C++0x for not coming already! Thank you for clarifying this. – LiraNuna Feb 15 '11 at 08:25
8

It's so sad that templatetypedefs are illegal. ;-) – templatetypedef Feb 15 '11 at 08:29
1

@templatetypedef: Then what are you doing here? ;) – Xeo Feb 15 '11 at 08:33
@templatetypedef : I knew you were illegal; thanks for admitting so :P – Nawaz Feb 15 '11 at 08:38

score 2 · Answer 2 · answered Feb 15 '11 at 08:42

2

Just for the sake of listing an alternative:

template <typename C>
struct test01 : test<0, 1, C> { };

test01<4> my_test014;

This does create new, distinct types and not simply aliases for instantiations of the base template :-(.

answered Feb 15 '11 at 08:42

Tony Delroy

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Sadly it doesn't guard against: `test<0, 1, 2>* x = new test01<2>(); delete x; // oops, UB!`. – GManNickG Feb 15 '11 at 08:53
@GMan: obviously true if you don't bother to put a `virtual` destructor in `test<>` (same old issue as derivation from Standard containers, and practically irrelevant in my experience and coding style, but everyone lives by the concerns that seem significant to them...) – Tony Delroy Feb 15 '11 at 09:24
Since I am doing mostly template meta-programming, this is fine by me. However, I do understand the difference. – LiraNuna Feb 15 '11 at 09:27
Does not work for #include template struct MyShared : public std::shared_ptr{}; – Cobaia Jun 30 '14 at 13:50
@Cobala: it I was inclined to put as much effort and detail into this comment as you did for yours, I'd just say "you're wrong", but to be more constructive ;-P... here's [an ideone example](http://ideone.com/7yN8hO) proving it "works", but more seriously you probably want to add a convenience function or two - e.g. at least constructor forwarding ala `using std::shared_ptr::shared_ptr;` inside `MyShared`. Cheers. – Tony Delroy Jun 30 '14 at 16:37

Template typedef of a template

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