Hi Eric, On Wed, Jun 22, 2011 at 11:05:33PM -0700, eric wrote: > Dear gcc programers: > > I tried to copy (from internet download book's example source code, > from > http://comscigate.com/BookCode/cppbooks.htm > on C++ Primer 3rd Ed, written by Lippman , Lajoie > that source code is written before 1999 and I guess is compiled based on > borland's compiler > I tried to compile/run chap17, text_query.C > attach 4 related files(I already modify a little bit, ie replace > iostream.h by iostream and add using namespace std;) > ------------------ > eric@eric-laptop:~/CppPrimer3/download/chap17$ g++ Query.C text_query.C > In file included from Query.C:1:0: > Query.h:27:41: error: type/value mismatch at argument 3 in template > parameter list for ‘template<class _Key, class _Compare, class _Alloc> > class std::set’ > Query.h:27:41: error: expected a type, got ‘allocator’ > Query.h:28:36: error: type/value mismatch at argument 2 in template > parameter list for ‘template<class _Tp, class _Alloc> class std::vector’ > Query.h:28:36: error: expected a type, got ‘allocator’ > ---------- > this is just first 4 errors of the whole(a lot). > /* which I believe is gcc specific(4.5.2), so I decide not post to > comp.lang.c++ */ No, it's a problem in your code. GCC is absolutely right. GCC states that in Query.h line 27 you're using "allocator" as a template parameter. At this position, you would need a type, but "allocator" is no type (but nevertheless: allocator is some object/word known to the compiler). The first line in question is the following: > const set< short,less<short>,allocator > *solution(); Due to your "using namespace std;" the word allocator is resolved to the standard template "std::allocator" defined in <memory>. However, std::allocator is a TEMPLATE, so you have to pass the type of variable which you want to allocate: > const set< short,less<short>,allocator<short> > *solution(); allocator<short> IS a type, so gcc will be happy :-) (and of course you need equivalent changes throughout your code). On the other hand: if you don't pass a third parameter to std::set, it will use by default std::allocator<T> (with T being the appropriate type). The same is true for the second template parameter: std::set defaults to use std::less<T> for comparisons. So, you can just replace this line by > const set< short > *solution(); which will give you the same behaviour. HTH, Axel
