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typedef iterator problem

Started by wanzi Apr 26, 2006 at 5:58 AM 4 replies 4.4k views

Original Post

Apr 26, 2006 at 5:58 AM

Ok, I've got the following code


template<typename T>
class Test
{
public:
	void Foo()
	{
		typedef std::vector<T>::iterator vec_itor;	// 1
	}

	typedef std::vector<T> vec_type;			// 2
	typedef std::vector<T>::iterator vec_itor;		// 3!!(Problem)
};

The compiler gives error for 3, saying "std::vector::iterator : dependent name is not a type". If I comment out 3, it runs properly. What puzzels me is that wht 1 and 2 work, but 3 doesn't. Aren't 1 and 3 basically the same??

Enigma

Apr 26, 2006 at 6:22 AM

The problem is known as "dependant name lookup", or "two-phase name lookup". Basically the compiler cannot know whether std::vector< T >::iterator is a type or a static member until it knows the type T (because of possible template specialisations). Therefore you must tell the compiler that std::vector< T >::iterator is a type by using the typename keyword:

typedef typename std::vector< T >::iterator vec_itor;.

I don't know why #1 works. I don't believe it should and gcc agrees with me. I'd advise adding typename into that line as well.

Σnigma

wanzi

Apr 26, 2006 at 6:32 AM

Thanks for the quick reply~~~~

For anyone interested in "Two-Phase Name Lookup", I just found a codeproject article http://www.codeproject.com/cpp/TwoPhaseLookup.asp

Emmanuel Deloget

Apr 26, 2006 at 6:38 AM

Enigma, doesn't #1 work because if Foo() is not called, it is not instantiated (and thus not compiled) - and if Foo() is called, vector is known (and vector::iterator is then known as a type)?

template <class T> class A{public:   typedef std::vector<T> vec_type;   typedef typename std::vector<T>::iterator iter_type;   void foo()   {      I can put whatever I want here.   }};

The code compiles correctly unless foo() is called.

Without this feature, metaprogramming using templates would be rather hard :)

Regards,

Polymorphic OOP

Apr 26, 2006 at 11:39 AM

Quote:
Original post by Emmanuel Deloget
Enigma, doesn't #1 work because if Foo() is not called, it is not instantiated (and thus not compiled) - and if Foo() is called, vector is known (and vector::iterator is then known as a type)?

While that may very well be happening with whatever compiler is being used, that is not correct behavior. The error should occur as soon as the template definition is encountered, not when it is being instantiated.

stylin

Apr 27, 2006 at 2:29 AM

Yes, the compiler should (does) try to validate all execution paths, even the ones that get optimized out. This is a good thing so we can take advantage of some of the more obscure metaprogramming techniques (like compile-time short-circuiting on 0-length array declarations).

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