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James Kuyper
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 Posted: Fri Jul 18, 2003 8:29 pm Post subject: Re: Article: allocators...exceptions...new/delete [reply to: |
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[email]dhruvbird (AT) gmx (DOT) net[/email] (Dhruv Matani) wrote in message news:<1058347888.3149.18.camel (AT) home (DOT) free>...
| Quote: | On Sat, 12 Jul 2003 09:24:06 -0400, Richard Smith wrote:
"Dhruv" <dhruvbird (AT) gmx (DOT) net> wrote in message
news:pan.2003.07.10.18.19.19.500436 (AT) gmx (DOT) net...
On Tue, 08 Jul 2003 02:32:22 +0000, Dhruv wrote:
....
If you want to do
template <class T> class my_allocator;
std::vector< int, my_allocator > v;
it is not OK.
I haven't quite understood what you're trying to say here, and why won't
the 2nd eg. work.
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The second template argument to any of the standard container classes
is a type, not a template name. my_allocator is a template.
my_allocator<T> for a given value of T, is a type.
....
| Quote: | Now, it would be great if this would work:
#include <memory
#include
#include
template //NOTE.
struct foo {
//type t;
some_alloc<type> ia;
//now, if you want a node allocator, just do:
//some_alloc<node_struct> na;
};
int main ()
{
foo <nstd::list_node_allocator f; //NOTE.
foo <boost::fast_pool_allocator f1; //NOTE.
}
I'm not too sure about whether the code is even correct. Please tell me
if
it's correct, because I don't trust g++ much on template issues.
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No, it's not. You've declared that the first template argument of
foo<> is a template name; you're not allowed to provide a type
argument instead.
| Quote: | Something that I'm quite sure about is that the boost allocator example
(f1 object thing) does not compile because of the the n number of
parameters that it has.
One more thing: Is it legal to access 'type' within the struct, as in,
can you do something like this inside the struct:
some_alloc<type> alloc;
It doesn't seem to work on g++.
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No, the 'type' is just a dummy parameter; what you're trying is
exactly analogous to the following:
void g(int (*f)(int i))
{
f(i);
}
Where g corresponds to foo, f corresponds to some_alloc, and i
corresponds to 'type'. The reason why you can't access 'i' is exactly
analogous to the reason you can't use 'type'.
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