I have a complex object that I want to be able to pass into a std::ostringstream
with the <<
operator just like a string or int. I want to give the ostringstream the object's unique id (int) and/or name (string). Is there an operator or method I can implement in my class to allow this to work?
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Mar
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5Yes just overload the << operator - see http://stackoverflow.com/questions/4421706/operator-overloading – Martin Beckett Jun 05 '12 at 22:01
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1unique id in what context? In an instance of running process you could use it's memory address. – Ivarpoiss Jun 05 '12 at 22:02
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They are "Student" objects, for a class project. Each Student object has a name, id, address, and phone number, and are built from a txt file. – Mar Jun 05 '12 at 22:09
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@MartinBeckett - overload << for my class, or for `ostringstream`? – Mar Jun 05 '12 at 22:10
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@mouseas - for your class. Each class simply has a function that is called when you ask for '<<' which prints the values. It's just that for int/float/double etc they are built-in – Martin Beckett Jun 05 '12 at 22:11
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Well, make an answer, then. What should the return be? – Mar Jun 05 '12 at 22:15
1 Answers
4
Define an operator overload in the same namespace as your class:
template<typename charT, typename traits>
std::basic_ostream<charT, traits> &
operator<< (std::basic_ostream<charT, traits> &lhs, Your_class const &rhs) {
return lhs << rhs.id() << ' ' << rhs.name();
}
If the output function needs access to private members of your class then you can define it as a friend function:
class Your_class {
int id;
string name;
template<typename charT, typename traits>
friend std::basic_ostream<charT, traits> &
operator<< (std::basic_ostream<charT, traits> &lhs, Your_class const &rhs) {
return lhs << rhs.id << ' ' << rhs.name;
}
};
Note that this does not result in a member function, it's just a convenient way to declare and define a friend function at once.

bames53
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1The `traits` template parameter has a default value, so it can be skipped. – AlwaysLearning Dec 09 '15 at 11:51