recursive type template

[n.torrey.pines]

I have the following tree definition:

template<typena me T>
struct tree {
T first;
vector<tree<T second; // branches
};

which compiles successfully. What I'd like to do though is to use
another template like 'pair', because I might have a bunch of useful
methods already defined in 'pair':

template<typena me T>
struct tree : public pair<T, tree<T {};

This doesn't compile. I tried some forward declarations with no
success. Is there any way to do it?

[Kai-Uwe Bux]

n.************@ gmail.com wrote:

I have the following tree definition:

template<typena me T>
struct tree {
T first;
vector<tree<T second; // branches
};

which compiles successfully.

Maybe, but the program has undefined behavior: standard templates may only
be instantiated with complete types. It may not show in the implementation
of std::vector in your library, but it might fail in others. The reason
that it is not guaranteed to fail is that the straight forward
implementation of std::vector only uses T* internally. Thus, you have to
put in an extra check to detect incomplete types. Some implementations of
std::vector do that.

What I'd like to do though is to use
another template like 'pair', because I might have a bunch of useful
methods already defined in 'pair':

template<typena me T>
struct tree : public pair<T, tree<T {};

This doesn't compile. I tried some forward declarations with no
success. Is there any way to do it?

Not really. You could do:

#include <utility>

template < typename T >
struct tree : public std::pair< T, tree<T>* {};

int main () {

tree<inta;

}
Best

Kai-Uwe Bux

[David W]

<n.************ @gmail.comwrote in message
news:11******** **************@ b28g2000cwb.goo glegroups.com.. .

I have the following tree definition:

template<typena me T>
struct tree {
T first;
vector<tree<T second; // branches
};

which compiles successfully. What I'd like to do though is to use
another template like 'pair', because I might have a bunch of useful
methods already defined in 'pair':

template<typena me T>
struct tree : public pair<T, tree<T {};

This doesn't compile. I tried some forward declarations with no
success. Is there any way to do it?

If your 'pair' is really std::pair, and that pair has an instance of each type passed to it as a
member (I assume it does), then you've effectively got struct tree having itself as a member, which
is impossible (since that member has another instance of itself as a member, ad infinitum), and in
its base class to boot, which is even more impossible (???). This problem isn't anything to do with
templates.

Uh, are you sure that 'tree' is a kind of 'pair'?

DW

[amparikh]

n.torrey.pi...@ gmail.com wrote:

I have the following tree definition:

template<typena me T>
struct tree {
T first;
vector<tree<T second; // branches
};

which compiles successfully. What I'd like to do though is to use
another template like 'pair', because I might have a bunch of useful
methods already defined in 'pair':

template<typena me T>
struct tree : public pair<T, tree<T {};

tree<Ttemplate class is not completely defined at the point at which
is being referred as the base class.

Not sure what you are trying to do, but try using CRTP.
Have some of the functionality you need in the base and then pass the
Derived as the templare parameter to the Base, to use it in the
std::pair that you define in the Base.

>
This doesn't compile. I tried some forward declarations with no
success. Is there any way to do it?

[shuisheng]

n.************@ gmail.com wrote:

I have the following tree definition:

template<typena me T>
struct tree {
T first;
vector<tree<T second; // branches
};

which compiles successfully.

This is because the vector contains the data pointers rather than the
data itself. So you can construct a instance of the tree class in the
memory since you know its size.

>
What I'd like to do though is to use
another template like 'pair', because I might have a bunch of useful
methods already defined in 'pair':

template<typena me T>
struct tree : public pair<T, tree<T {};

This doesn't compile. I tried some forward declarations with no
success. Is there any way to do it?

But here the compiler don't know the size of the class because the tree
in pair<T, tree<T is not define well when it is used to construct
the derived tree. So the compiler give an error.

Try this:

template<typena me T>
struct tree : public pair<T, tree<T>* {};

Bests,

Shuisheng

[Gernot Frisch]

>I have the following tree definition:

>
template<typena me T>
struct tree {
T first;
vector<tree<T second; // branches
};

maybe this way:

template<typena me Tstruct tree
{
T first;
std::vector<tre e<T subtrees;
};
int main()
{
tree<pair<int, std::string mytree;

}

[Kai-Uwe Bux]

Gernot Frisch wrote:

>>I have the following tree definition:

template<typen ame T>
struct tree {
T first;
vector<tree<T second; // branches
};

maybe this way:

template<typena me Tstruct tree
{
T first;
std::vector<tre e<T subtrees;
};

That is UB according to [17.4.3.6/2] item 5:

In particular, the effects are undefined in the following cases:
...
? if an incomplete type (3.9) is used as a template argument when
instantiating a template component.

>
int main()
{
tree<pair<int, std::string mytree;

}

Best

Kai-Uwe Bux

[Gernot Frisch]

That is UB according to [17.4.3.6/2] item 5:

In particular, the effects are undefined in the following cases:
...
? if an incomplete type (3.9) is used as a template argument when
instantiating a template component.

and if I substitute "typename" by "class"?

[shuisheng]

n.************@ gmail.com wrote:

I have the following tree definition:

template<typena me T>
struct tree {
T first;
vector<tree<T second; // branches
};

which compiles successfully. What I'd like to do though is to use
another template like 'pair', because I might have a bunch of useful
methods already defined in 'pair':

template<typena me T>
struct tree : public pair<T, tree<T {};

This doesn't compile. I tried some forward declarations with no
success. Is there any way to do it?

template<typena me T>
struct tree : public pair<T, tree<T>* {};

make the second part of pair to be a pointer.t

[Kai-Uwe Bux]

Gernot Frisch wrote:

>

>That is UB according to [17.4.3.6/2] item 5:

In particular, the effects are undefined in the following cases:
...
? if an incomplete type (3.9) is used as a template argument when
instantiating a template component.

and if I substitute "typename" by "class"?

The word "typename" does not occur anywhere in what you quoted. So, what
exactly do you mean?
Best

Kai-Uwe Bux