关于模板函数中的矢量迭代器的问题

Question

I'm trying to learn the STL library and I'm having a weird problem. This code compiles perfectly:

void Show(vector<int> myvec)
{
    vector<int>::iterator it;
    cout << "Vector contains:";
    for( it = myvec.begin(); it < myvec.end(); it++) 
    {
         cout << " " << *it;
    }
    cout << endl;
}

while this one gives me an error message at compile time:

template <class T> 
void Show2(vector<T> myvec)
{
    vector<T>::iterator it;
    cout << "Vector contains:";
    for( it = myvec.begin(); it < myvec.end(); it++)
    {
         cout << " " << *it;
    }
    cout << endl;
}

The error is:

$ g++ hello.cpp
hello.cpp: In function ‘void Show2(std::vector<T, std::allocator<_Tp1> >)’:
hello.cpp:19: error: expected ‘;’ before ‘it’
hello.cpp:21: error: ‘it’ was not declared in this scope

It seems a very simple mistake, but I couldn't find it.

Answer 1

You need to say typename vector<T>::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<T> const &myvec) and using const_iterator would be wiser.

Answer 2

You need this:

typename vector<T>::iterator it;

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

Answer 3

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<T>::iterator it;

Answer 4

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

Answer 5

In the first instance the parameter, although it uses a template, is not a template, it is a fully defined class ( vector<int> )

In the latter instance the parameter is a template on type T and thus requires typename