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Neal D. Becker
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 Posted: Thu Jul 22, 2004 10:42 pm Post subject: typedefs and references |
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This simple example doesn't compile:
#include <vector>
template<typename cont_t>
void noop (cont_t in) {
for (typename cont_t::iterator i = in.begin(); i != in.end(); i++) ;
}
void F() {
std::vector<int> v;
noop<std::vector (v);
}
g++34 -c Test.cc
Test.cc: In function `void noop(cont_t) [with cont_t = std::vector<int,
std::allocator&]':
Test.cc:10: instantiated from here
Test.cc:5: error: `std::vector<int, std::allocator&' is not a class,
struct, or union type
The problem can be solved using boost::remove_reference. This is tedious.
Is it reasonable that this code should be rejected? If a type T has a
typedef, then would T& not also have that typedef?
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Ali Cehreli
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| Posted: Fri Jul 23, 2004 2:12 am Post subject: Re: typedefs and references |
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On Thu, 22 Jul 2004 15:42:54 -0700, Neal D. Becker wrote:
| Quote: | This simple example doesn't compile:
#include <vector
template
void noop (cont_t in) {
for (typename cont_t::iterator i = in.begin(); i != in.end(); i++) ;
}
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I think you want to define noop as taking a reference. Containers are
generally expensive to copy. Additionally, if there is no change on
the container, it better be const-reference:
template
void noop(cont_t const & in);
| Quote: |
void F() {
std::vector<int> v;
noop<std::vector (v);
}
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I think you are putting the '&' at the wrong place. The type that noop
will use is std::vector<int>. How an object of that type is passed to
noop is defined in the function signature.
I think that making the 'cont_t const & in' change above is all you
need in this case.
| Quote: | g++34 -c Test.cc
Test.cc: In function `void noop(cont_t) [with cont_t = std::vector<int,
std::allocator&]':
Test.cc:10: instantiated from here
Test.cc:5: error: `std::vector<int, std::allocator&' is not a
class, struct, or union type
The problem can be solved using boost::remove_reference. This is
tedious. Is it reasonable that this code should be rejected? If a type
T has a typedef, then would T& not also have that typedef?
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typedef just renames an existing type. A type can't "have a typedef",
but may "have an alias through the use of typedef."
I see T& as a reference to T; that's all. Maybe I am misunderstanding
your question...
Ali
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Dieter Beaven
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| Posted: Fri Jul 23, 2004 9:53 am Post subject: Re: typedefs and references |
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Hi Neal,
Looking at the iterator declaration in noop's for loop, I think the
compiler would interpret it as the following:
cont_t::iterator => std::vector<int, std::allocator&::iterator
i.e. the statement is asking for the iterator_type belonging to a
reference, not the iterator type of a std::vector<int,
std::allocator.
One solution would be to declare void noop( cont_t& in ), and dispense
with the reference in the template instantation. Else maybe an
appropriate typedef within noop() to specify to the compiler exactly
what you intend.
I would favour the former unless you have a good reason to want to
declare template instanciations of references. Things like
std::vector<int&> tend not to compile due to reference to reference
issues; which aren't currently supported in C++. If you need to
support references in template instanciations Alexandrescu's Modern
C++ Design has some useful techniques.
Regards,
Dieter
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