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E.D.
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 Posted: Thu Dec 30, 2004 8:16 pm Post subject: Why doesn't this compile? |
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Hi!
In the following code I get a compile error when I call f1(). I use
g++ 3.2.2. Is this a compiler problem and if not where does this code
violate the standard?
Thanx in advance,
ED.
#include <set>
template <typename T>
struct F {
typedef std::set<F> Set;
};
class C {
public:
template<typename T>
void f1(const typename F<T>::Set& s) {
}
template<typename T>
void f2(const std::set<T>& s) {
}
};
int main(int argc, char** argv) {
std::set<int> s;
C c;
c.f1(s); // compile error
c.f1<int>(s); // also does not compile
c.f2(s); // works
return 0;
}
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Ralph Zhang
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| Posted: Fri Dec 31, 2004 1:14 pm Post subject: Re: Why doesn't this compile? |
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You probably mean
template <typename T>
struct F {
typedef std::set<T> Set;
// ^
};
And then
c.f1<int>(s);
compiles.
As for
c.f1(s);
because the compiler cannot deduce the template parameter, it won't
compile at all.
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L.Suresh
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| Posted: Fri Dec 31, 2004 1:18 pm Post subject: Re: Why doesn't this compile? |
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Deduction fails because its a non deduced context. Qualified type names
is a non deduced context.
14.8.2.4 Deducing template arguments from a type
"
If a template parameter is used only in nondeduced contexts and is not
explicitly specified, template argument deduction fails.
The nondeduced contexts are:
- The nested-name-specifier of a type that was specified using a
qualified-id.
- A type that is a template-id in which one or more of the
template-arguments is an expression that references a
template-parameter.
"
--lsu
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Victor Bazarov
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| Posted: Fri Dec 31, 2004 1:22 pm Post subject: Re: Why doesn't this compile? |
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E.D. wrote:
| Quote: | In the following code I get a compile error when I call f1(). I use
g++ 3.2.2. Is this a compiler problem and if not where does this code
violate the standard?
Thanx in advance,
ED.
#include <set
template
struct F {
typedef std::set
|
Did you mean to write
typedef std::set<T> Set;
? 'F' is not a type-id here, it's a template-id, so who knows what
the compiler is trying to do. And 'std::set' needs a type-id as its
first argument.
| Quote: | };
class C {
public:
template<typename T
void f1(const typename F
}
template<typename T
void f2(const std::set
}
|
It seems that if you wanted 'f1' and 'f2' to have the same argument types,
then you definitely wanted F<T>::Set to be std::set<T>, not std::set<F>
| Quote: | };
int main(int argc, char** argv) {
std::set<int> s;
C c;
c.f1(s); // compile error
|
Will still fail to compile because it can't figure out 'T', since 'Set'
can be defined differently in specific instantiations of 'F'. 'Set' is
a dependent name, and there are special rules that allow the compiler
not to try to figure out T in those cases, IIRC.
| Quote: | c.f1<int>(s); // also does not compile
|
Will compile when you make the change as I suggested.
| Quote: | c.f2(s); // works
return 0;
}
|
V
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