No match for 'operator

问题

I am developing gsoap web service where I am retrieving vectors of objects in return of a query. I have two ways to do it: first by simple loop and by iterator. None of them working.

The error is:

error: no match for 'operator<<' in 'std::cout mPer.MultiplePersons::info.std::vector<_Tp, _Alloc>::at<PersonInfo, std::allocator<PersonInfo> >(((std::vector<PersonInfo>::size_type)i))'

MultiplePersons mPer; // Multiple Person is a class, containing vector<PersonInfo> info
std::vector<PersonInfo>info; // PersonInfo is class having attributes Name, sex, etc.
std::vector<PersonInfo>::iterator it;

cout << "First Name: \t";
cin >> firstname;
if (p.idenGetFirstName(firstname, &mPer) == SOAP_OK) {
    // for (int i = 0; i < mPer.info.size(); i++) {
    //    cout << mPer.info.at(i); //Error
    //}
    for (it = info.begin(); it != info.end(); ++it) {
        cout << *it; // Error
    }

} else p.soap_stream_fault(std::cerr);

}

It's obvious that operator overloading operator<< in cout is the problem. I have looked at several problems related to this, but no one helped me out. If someone can provide a concrete example on how to solve it, it would be very appreciated. (Please do not talk in general about it, I am new to C++ and I have spent three days on it searching for solution.)

回答1:

You need to provide an output stream operator for PersonInfo. Something like this:

struct PersonInfo
{
  int age;
  std::string name;
};

#include <iostream>
std::ostream& operator<<(std::ostream& o, const PersonInfo& p)
{
  return o << p.name << " " << p.age;
}

This operator allows expressions of the type A << B, where A is an std::ostream instance (of which std::cout is one) and B is a PersonInfo instance.

This allows you do do something like this:

#include <iostream>
#include <fstream>
int main()
{
  PersonInfo p = ....;
  std::cout << p << std::endl; // prints name and age to stdout

  // std::ofstream is also an std::ostream, 
  // so we can write PersonInfos to a file
  std::ofstream person_file("persons.txt");
  person_file << p << std::endl;
}

which in turn allows you to print the de-referenced iterator.

回答2:

The result of *it is an L-value of type PersonInfo. The compiler is complaining that there is no operator<< which takes a right-hand side argument of type PersonInfo.

For the code to work, you need to provide such an operator, for example like this:

std::ostream& operator<< (std::ostream &str, const PersonInfo &p)
{
  str << "Name: " << p.name << "\nAge: " << p.age << '\n';
  return str;
}

The exact implementation of the operator depends on your needs for representing the class in output, of course.

回答3:

What it's telling you is that there isn't a known wway to cout (console output) the contents of *it.

it is an iterator - think of this like a pointer in a list

the list is info so *it is current item in the info, which is a list of PersonInfo items.

So cout << *it; says output to the console the PersonInfo that it is currently referencing.

But the error message is telling you that the compiler doens't know how PersonInfo should be rendered to the console.

What you need to do is create an operator called << that takes an object that cout is (ostream) and a PersonInfo object and then writes the various bits of the PersonInfo to cout.

来源：https://stackoverflow.com/questions/15449508/no-match-for-operator-in-stdcout

标签

c++

cout

No match for 'operator<<' in std::cout [duplicate]

问题

回答1:

回答2:

回答3: