Question about vector iterator in template functions?

You need to say typename vector::iterator it On another note, you're passing vector s by value. That means the entire vector gets copied in the function call void Show(vector const &myvec) and using const_iterator would be wiser.

You need to say typename vector::iterator it. On another note, you're passing vectors by value. That means the entire vector gets copied in the function call.

Void Show(vector const &myvec) and using const_iterator would be wiser.

2 If he passes a const ref, you need to also tell him to use a const_iterator. – Erik Mar 4 at 11:08 @Erik: absolutely right, updated my post. – larsmans Mar 4 at 11:09 thanks for the tip.

– Rafael S. Calsaverini Mar 4 at 11:22.

You need this: typename vector::iterator it; This tells the compiler that vector::iterator should be treated as a type, something it can't assume since iterator is dependent on what T is.

Some compilers have problems detecting what is a member name and what is a type name, when inside templates. Try writing something like this in the first line of your template function body. Typename vector::iterator it.

1 Correct, but it is not really a compiler problem. The C++ Standard mandates that the compiler treat a dependent name in a template as a member and not a type, unless typename is used. – decltype Mar 4 at 11:11 1 It's not "some compilers".

The grammar simply does not allow it. "Some compilers" can double-guess programmer by retrying with typename if parsing fails without it and thus compiling some incorrect cases. – Jan Hudec Mar 4 at 11:18 You are right.

Point taken! – CygnusX1 Mar 4 at 11:29.

Maybe it works using typename vector::iterator it; Your compiler cannot know that there is an inner class iterator.

In the first instance the parameter, although it uses a template, is not a template, it is a fully defined class (vector) In the latter instance the parameter is a template on type T and thus requires typename.

You need to say typename vector::iterator it .

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