No match for operator

问题

I'm trying to print every element inside a vector like this:

vector<users>::iterator i;

for(i = userlist.begin(); i<userlist.end(); i++)
{
        cout << *i << "\n";
}

Then I'm getting an error like this:

no match for 'operator<<' in 'std::cout << (&i)->__gnu_cxx::__normal_iterator<_Iterator, _Container>::operator* [with _Iterator = users*, _Container = std::vector<users, std::allocator<users> >]()'

Is it anything obvious I've missed?

回答1:

Have you defined a function with this signature?:

std::ostream & operator<<(std::ostream &, const users &);

It should not be a member function of users, although it may or may not be a friend, up to you. The prototype should go in the header file of class users, and the body should go in the source(.cpp) file. I have no idea how your users class is defined, or how you would want to format the output, but the function definition should look something like this:

std::ostream & operator<<(std::ostream & os, const users & U)
{
    os << U.some_data_members;
    os << U.and_or_some_member_functions();
    os << whatever;
    return os;
}

回答2:

Once you've defined std::ostream &operator<<(std::ostream &, user&);, consider changing your code to use std::copy instead of a for loop:

// leaving off the `std::`, you're not using it for `cout`.
// 
copy(userlist.begin(), userlist.end(), ostream_iterator<user>(cout, "\n"));

回答3:

Have you defined stream operator for the users class? If not, do so.

回答4:

You need to write an overload of ostream::operator<<() that takes a an instance of users, or write some conversion operator that will provide an auto-conversion from user to some type that one of the operator<<() versions knows about.

回答5:

You need to define your own public function operator<< taking parameters of an ostream and a users:

std::ostream& operator<<(std::ostream&, users&);

Sorry, is it users or user?

来源：https://stackoverflow.com/questions/5707827/no-match-for-operator-in-stdcout

标签

c++

vector

cout