I'm trying to write a function to print a representation of common STL containers (vector, list, etc..). I gave the function a template parameter T which, for example, might represent vector. I'm having problems getting an iterator of type T.
vector<int> v(10, 0);
repr< vector<int> >(v);
...
template <typename T>
void repr(const T & v)
{
cout << "[";
if (!v.empty())
{
cout << ' ';
T::iterator i;
for (i = v.begin();
i != v.end()-1;
++i)
{
cout << *i << ", ";
}
cout << *(++i) << ' ';
}
cout << "]\n";
}
...
brett@brett-laptop:~/Desktop/stl$ g++ -Wall main.cpp
main.cpp: In function ‘void repr(const T&)’:
main.cpp:13: error: expected ‘;’ before ‘i’
main.cpp:14: error: ‘i’ was not declared in this scope
main.cpp: In function ‘void repr(const T&) [with T = std::vector<int, std::allocator<int> >]’:
main.cpp:33: instantiated from here
main.cpp:13: error: dependent-name ‘T::iterator’ is parsed as a non-type, but instantiation yields a type
main.cpp:13: note: say ‘typename T::iterator’ if a type is meant
I tried 'typename T::iterator' as the compiler suggested, but only got a more cryptic error.
Edit: Thanks for the help guys! Here's a working version for anyone who wants to use this function:
template <typename T>
void repr(const T & v)
{
cout << "[";
if (!v.empty())
{
cout << ' ';
typename T::const_iterator i;
for (i = v.begin();
i != v.end();
++i)
{
if (i != v.begin())
{
cout << ", ";
}
cout << *i;
}
cout << ' ';
}
cout << "]\n";
}
You need typename
to tell the compiler that ::iterator
is supposed to be a type. The compiler doesn't know that it's a type because it doesn't know what T is until you instantiate the template. It could also refer to some static data member, for example. That's your first error.
Your second error is that v
is a reference-to-const. So, instead of ::iterator
you have to use ::const_iterator
. You can't ask a constant container for a non-const iterator.