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.
Related
I have this code
template <typename T>
class KeyValueProperty {
protected:
T value = T();
std::string key = "";
public:
KeyValueProperty(const std::string & key) : key(key) { }
T & operator = (const T &i) { return value = i; };
operator const T & (){ return value; };
};
struct T2 {
KeyValueProperty<std::string> x {"x"};
KeyValueProperty<double> y {"y"};
};
and in main
T2 tx;
tx.x = "hellow";
tx.y = 10;
std::cout << static_cast<std::string>(tx.x) << ::std::endl;
std::cout << tx.y << ::std::endl;
This is working correctly. However, doing only this
std::cout << tx.x << ::std::endl;
will end up in
error C2679: binary '<<': no operator found which takes a right-hand operand of type 'Test::KeyValueProperty' (or
there is no acceptable conversion)
Is it possible to have auto-conversion, or I must manually call casting?
The reason t.y works even without a custom operator<< is because there already exists an operator<<(std::ostream&, double), and the compiler can also see that it can make a double out of your class. It does so and we're happy.
However, there is no operator<<(std::ostream&, std::string). If there was, the same logic would apply and we'd still be happy. Instead, there is:
template <class CharT, class Traits, class Allocator>
std::basic_ostream<CharT, Traits>&
operator<<(std::basic_ostream<CharT, Traits>& os,
const std::basic_string<CharT, Traits, Allocator>& str);
That is, a generic insertion operator for any kind of basic_string.
Although there exist some template arguments that would make this as if it were operator<<(std::ostream&, std::string), the compiler isn't going to try and guess which possible template arguments would allow it to subsequently match a conversion of your class to the result. There are too many combinations, so this isn't permitted.
This is why you had to explicitly turn your object into a std::string (aka std::basic_string<char>) - this removes one layer from the problem and it can do regular old type deduction to make this work.
The right solution is to give your wrapper class an insertion operator, to sidestep this issue.
You must provide an appropriate overload of operator<<, for example:
template<class T>
std::ostream& operator<<(std::ostream& os, KeyValueProperty<T> const& kvp)
{
return os << T(kvp);
}
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 am trying to overload the << operator, but I get the following error:
error: ambiguous overload for 'operator<<' in 'std::cout << "Test "'
..Followed by 5 billion other errors similar to:
c:\mingw\bin../lib/gcc/mingw32/4.5.2/include/c++/ostream:165:7: note:
candidates are: ...
This comes up because I'm using cout in my main.cpp file.
Here is my code:
In BinTree.h:
template <typename T>
class BinTree{
...
friend std::ostream& operator<< <>(std::ostream&, const T&);
In BinTree.cpp:
template <typename T>
std::ostream& operator<< (std:: ostream& o, const T& value){
return o << value;
}
Thanks in advance for any help you can give.
Your function has the same signature than the one already defined. This is why the compiler moans about ambigous overload. Your function tries to define a function to stream everything to a ostream. This function already exists in the standards library.
template <typename T>
std::ostream& operator<< (std:: ostream& o, const T& value){
return o << value;
}
What you perhaps want to do is write a function that defines how a BinTree is streamed (to everything). Please note that the stream type is templated. So if you chain the calls to the stream operator it streams the concrete type.
template <typename T, typename U>
T& operator<< (T& o, const BinTree<U>& value){
//Stream all the nodes in your tree....
return o;
}
Did you mean..
template<class T>
ostream& operator<<(ostream& os, const BinTree<T>& v){
typename BinTree<T>::iterator it;
for(it = v.begin(); it != v.end(); ++it){
os << *it << endl;
}
return os;
}
Post more code, till then see this part:
template <typename T>
std::ostream& operator<< (std:: ostream& o, const T& value){
return o << value;
}
This is doing nothing but calling itself. It is recursive call. The operator<< is defined to output value of type T, and when you write o<<value, it calls itself, as the type of value is T.
Second, since this is function-template, the definition should be provided in the .h file, not in .cpp file if you expect your code working by including .h file.
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.
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.)