how can i overload "<<" operator (for cout) so i could do "cout" to a class k
The canonical implementation of the output operator for any type T is this:
std::ostream& operator<<(std::ostream& os, const T& obj)
{
os << obj.get_data1() << get_data2();
return os;
}
Note that output stream operators commonly are not member functions. (That's because for binary operators to be member functions they have to be members of their left-hand argument's type. That's a stream, however, and not your own type. There is the exception of a few overloads of operator<<() for some built-ins, which are members of the output stream class.)
Therefor, if not all data of T is publicly accessible, this operator has to be a friend of T
class T {
friend std::ostream& operator<<(std::ostream&, const T&);
// ...
};
or the operator calls a public function which does the streaming:
class T {
public:
void write_to_stream(std::ostream&);
// ...
};
std::ostream& operator<<(std::ostream& os, const T& obj)
{
obj.write_to_stream(os);
return os;
}
The advantage of the latter is that the write_to_stream() member function can be virtual (and pure), allowing polymorphic classes to be streamed.
If you want to be fancy and support all kinds of streams, you'd have to templatize that:
template< typename TCh, typename TTr >
std::basic_ostream<TCh,TTr>& operator<<(std::basic_ostream<TCh,TTr>& os, const T& obj)
{
os << obj.get_data1() << get_data2();
return os;
}
(Templates, however, don't work with virtual functions.)
Related
I have some Logging::Logger class with the following functions:
template<typename T>
const Logger& Logger::operator<<(const T& in) const {
// ...
return *this;
}
const Logger& Logger::operator<<(std::ostream& (*os)(std::ostream&)) {
// ...
return *this;
}
And the following code:
loggerInstance << "ID: " << 5 << endl;
And I'm getting the following error though all operators seems to be implemented:
error C2678: binary '<<': no operator found which takes a left-hand
operand of type 'const Logging::Logger' (or there is no acceptable
conversion)
Of course, without endl everything is working.
I've looked at the following answer:
std::endl is of unknown type when overloading operator<<
What am I missing?
Because your overloaded operators return a const Logger &, it follows that they must be const class methods, in order for you to be able to chain them together:
const Logger& Logger::operator<<(std::ostream& (*os)(std::ostream&)) const
However, it's better if they were not const class members, and returned a Logger &, instead:
template<typename T> Logger& Logger::operator<<(const T& in)
Logger& Logger::operator<<(std::ostream& (*os)(std::ostream&))
This would be because, presumably, operator<< would be modifying the Logger instance, in some way. If not, you can use const objects and methods, here.
Im trying to build a vector of type template but keep getting an error
template<class T>
struct s{
T val;
public:
s(T a)
{
val = a;
};
friend ostream& operator<<(ostream& os, const T& a);
};
template <typename T>
ostream& operator<< (ostream& os, const T& a){
return os << a.val;
}
int main()
{
s <double> names(4) ;
s <int> figure(7);
s <string> word();
s <vector<int>*> arrays();
cout << names << endl;
cout << figure << endl;
cout << arrays << endl;
return 0;
}
I keep receiving the same error -
request for member 'val' in 'a', which is of non-class type 's<std::vector<int>*>()'|
Any advice will be greatly appreciated
s <vector<int>*> arrays();
Declares a function named arrays taking no arguments and returning a s<vector<int>*>. Remove the redundant parentheses or replace them with {} for C++11. Of course, your s doesn't support default construction, so you need to give it a default constructor - which can be as simple as giving the constructor a default argument:
s(T a = T())
{
val = a;
}
and it's better to use the member initialization list instead of assignment and take a const ref instead of by value, since T can be large:
s(const T& a = T()) : val(a) { }
In C++11, you should take by value and move a into val instead:
s(T a = T()) : val(std::move(a)) { }
Also,
template <typename T>
ostream& operator<< (ostream& os, const T& a){
return os << a.val;
}
would match everything under the sun. Better to match s only:
template <typename T>
ostream& operator<< (ostream& os, const s<T>& a){
return os << a.val;
}
Finally,
friend ostream& operator<<(ostream& os, const T& a);
befriends (and declares) a simple function rather than a function template. To befriend the template operator << as modified above:
template <typename U>
friend ostream& operator<<(ostream& os, const s<U>& a);
Or even better, remove the template operator and define the friend function inline (inside the class definition) instead:
friend ostream& operator<<(ostream& os, const s<T>& a){
return os << a.val;
}
This is more compact, and also limits friendship to the matching version of operator <<, rather than all instantiations of the template.
Demo.
s <vector<int>*> arrays();
This does not instantiate an object, it declares a function that returns s<vector<int>*>, and as a consequence:
cout << arrays << endl;
Tries to print out a function pointer by instantiating your templated operator<<:
template <typename T>
ostream& operator<< (ostream& os, const T& a){
return os << a.val;
}
with:
T = s<vector<int>*>(*)();
and so, a pointer to function a does not have the .val field which triggers the error.
The whole problem is that this
ClassName InstanceName();
is not creating an instance of ClassName using default constructor, but declares a function. In order to do what you wanted (and I presume that is to create instances with default constructors) use syntax
ClassName InstanceName;
So, in order to fix your errors, change
s <string> word();
s <vector<int>*> arrays();
to
s <string> word;
s <vector<int>*> arrays;
and add a default constructor to your class s.
Im trying to do an override on a simple struct like this:
struct Node {
int data1;
int data2;
ostream& operator<<(ostream &o, const Node &n)
{
o << "(a: " << data1 << ", b: " << data2 << ")";
return o;
}
};
Im getting: error C2804: 'operator <<' too many parameters
So, if i remove the second parameter:
ostream& operator<<(ostream &o)
Then i get: Error: binary '<<' : no operator found which takes a right-hand operand of type 'const Node'
What is coming on here?
operator<< needs two operands. There are two ways to define an operator<< overload function between two types.
As a member function. That's how operator<< overloads between std::basic_ostream and some of the basic types are defined. When you use std:cout << 10;, it gets resolved to the overload operator std::basic_ostream<char>::operator<<(int)
When you define it as member function, the LHS of the operator is an instance of the class. The RHS of the operator is the argument to the function. That's why when you define it as member function, it can only have one argument.
As a free function. That's how operator<< overloads between std::basic_ostream and custom types are defined. These functions must have two arguments. The first argument is the LHS of the operator and the second argument is the RHS of the operator.
For this reasons, you have to define operator<< overload between std::ostream and your class as a free function, with std::ostream& as the first argument type and Node const& as the second argument type.
std::ostream& operator<<(std:ostream& os, Node const& node);
At times like these, I wish the standard library had implemented a function:
template <typename T>
std::ostream& operator<<(std::ostream& os, T const& t)
{
return t.serialize(os);
}
Then, any class could provide the function
std::ostream& serialize(std::ostream& os) const;
and be ready to be used like all the basic types with std::ostream.
std::ostream& operator<<(std::ostream&, ...) needs to be a free function.
Move it outside the class and it will work.
The reason it is so is because defining operator<<(std::ostream&) (or other binary operators) inside the class implies that object is the LHS operand. You would have to write smth crazy like:
Node << std::cout;
Ambiguous overload for operator<<() is called when I add the overload function below
template <typename Container> ostream& operator<<(ostream& os, const Container& c)
{
copy(c.begin(), c.end(), ostream_iterator<typename Container::value_type>(os, " "));
return os;
}
the error is called on this function where it uses the <<.
void print_list(const list<int>& list_int)
{
for (list<int>::const_iterator it = list_int.begin(); it != list_int.end(); it++) cout << *it << " ";
}
(For reference, if anyone else is looking: http://ideone.com/YlX7q )
Your definition of operator<< can be instantiated as ::operator<<<int>(std::ostream&, const int&); this is ambigious with std::operator<<(std::ostream&, int). Calling the name of the type Container doesn't mean it is a container; overload resolution is done before the definition is instantiated.
Yes of course this cannot work.
You are introducing a templated overload, and the compiler don't know anymore what to use when you use that operator.
Simply you cannot do that.
You can do something like this:
template<class T>
friend ostream& operator<<(ostream& os, const MyClass<T>& r);
but you cannot do
template<class T>
friend ostream& operator<<(ostream& os, const T& r);
I have a class:
class foo {
private:
std::string data;
public:
foo &append(const char* str, size_t n) { data.append(str,n); }
// for debug output
template <typename T>
friend T& operator<< (T &out, foo const &f);
// some other stuff
};
template <typename T>
T& operator<< (T &out, foo const &f) {
return out << f.data;
}
I want this to work with any class that provides the << operator.
This works fine with std::cout as in:
std::cout << fooObject;
But the following fails:
BOOST_AUTO_TEST_CASE( foo_append_and_output_operator )
{
// fooObject is accessable here
const char* str = "hello";
fooObject.append(str, strlen(str));
output_test_stream output;
output << fooObject;
BOOST_CHECK( output.is_equal(str) );
}
g++ tells me that:
In function ‘T& operator<<(T&, const foo&)
[with T = boost::test_tools::output_test_stream]’:
error: invalid initialization of reference of type
‘boost::test_tools::output_test_stream&’ from expression of type
‘std::basic_ostream<char, std::char_traits<char> >’
What's going on?
I'm using Boost 1.34.1 on Ubuntu 8.04.
So I think I have an explanation, but no solution yet. output_test_stream implements its stream functionality by subclassing wrap_stringstream. The insertion-operator for this is a free function-template that looks like this:
template <typename CharT, typename T>
inline basic_wrap_stringstream<CharT>&
operator<<( basic_wrap_stringstream<CharT>& targ, T const& t )
{
targ.stream() << t;
return targ;
}
// ... further down in the same header
typedef basic_wrap_stringstream<char> wrap_stringstream;
Your operator is called with output_test_stream as the stream-type, and that makes this it's return-type. Your operator then calls the above operator, and just propagates the return value. The return value of the above operator however is a superclass of the returntype of your operator. When compiler tries to create the reference you want to return, it chokes, because it cannot initialize a reference to a subclass from a reference to a superclass, even if both refer to the same object. That make any sense?
You may know that already, but using output_test_stream as an std::ostream works:
class foo {
// [...]
friend
std::ostream& operator<< ( std::ostream &os, const foo &f );
};
std::ostream& operator<< ( std::ostream &os, const foo &f ) {
return os << f.data;
}
Is it a typo? You wrote
foo.append(str, strlen(str));
but foo is the name of the class and not an object.