Keep getting "no match for 'operator<<' C++ - c++

Question

I'm writing a program to read text from a file and I need to print out the position that each of the numbers appear in the file. For example, the .txt file looks like this:
one
two one two
three three two one
and my output should look like:
one: 0, 2, 7
two: 1, 3, 6
three: 4, 5
Everything works fine until I try to display the map of type string, list(int), then I get the whole ""no match for 'operator<<'" error.
Here is my code, any help would be greatly appreciated, thanks!
#include <iostream>
#include <fstream>
#include <string>
#include <map>
#include <list>
#include <vector>
using namespace std;
int main()
{
ifstream is("input.txt");
if (is.fail())
{
cout << "File I/O error!" << endl;
exit(1);
}
map<string, list<int> > m;
string word;
vector<string> v;
list<int> l, x, y;
while (is >> word)
{
v.push_back(word);
}
for (unsigned int i = 0; i < v.size(); ++i)
{
if (v[i] == "one")
l.push_back(i);
else if (v[i] == "two")
x.push_back(i);
else if (v[i] == "three")
y.push_back(i);
}
m["One"] = l;
m["Two"] = x;
m["Three"] = y;
for (map<string, list<int> >::iterator i = m.begin(); i != m.end(); ++i)
cout << (*i).first << ", " << (*i).second << endl;
return 0;
}

Answer 1

The problem is when you try to output (*i).second. As i is of type map<string, list<int> >::iterator (*i).second is a list<int> and c++ does not know how to output it.
You have two options - either overload ostream& operator<<(ostream& out, const list<int>& l) or use an inner cycle to output the elements one by one. I personally recommend the second option as overloading the operator for such "popular" type may be dangerous.

Answer 2

The simplest solution:
// !!! DO NOT DO THIS AT HOME OR AT ALL !!!
namespace std {
template <typename T, typename A>
ostream& operator<<(ostream& out, list<T, A> const& l) {
if (l.empty()) { return out << "[]"; }
out << '[';
bool first = true;
for (auto const& t: l) {
if (first) { first = false; } else { out << ", "; }
out << t;
}
return out << ']';
} // operator<<
} // namespace std
Unfortunately... it is strictly forbidden to do so (you are only allowed to specialize existing templates in the std namespace, not add any overload).
Therefore, the best standard compliant solution is to:
Declare a new stream of your own (which forwards everything to a std::ostream&)
Overload this operator for your new stream (in its namespace)
Only ever use this new stream afterward
Feel free to bitch at the clumsiness...
Example of custom stream to get you started:
namespace project {
class OStream {
public:
explicit OStream(std::ostream& out): _out(out) {}
template <typename T>
OStream& operator<<(T const& t) { print(*this, t); return *this; }
template <typename T>
void push(T const& t) { _out << t; }
private:
std::ostream& _out;
}; // class OStream
// Generic Operator (directly forwards to `std::ostream`)
template <typename T>
void print(OStream& out, T const& t) { out.push(t); }
// STL Containers
template <typename It>
void print_range(OStream& out, It begin, It end) {
if (begin == end) { out << "[]"; return; }
out << '[' << *begin;
for (++begin; begin != end; ++begin) { out << ", " << *begin; }
out << ']';
} // push_range
template <typename T, typename A>
void print(OStream& out, std::list<T, A> const& l) {
print_range(out, l.begin(), l.end());
} // operator<<
} // namespace project
// usage
int main() {
std::list<int> example = { 1, 2, 3, 4 };
project::OStream(std::cout) << example << "\n";
}
Note: there are other solutions than creating a custom stream, such as copy pasting std::copy(example.begin(), example.end(), std::ostream_iterator<int>(std::cout, ", ")); everywhere you need to print a list, but I am yet to meet another handy one.