Why the line where i am trying to print the "reference_wrapper for string" is giving error for unsupported operator<< for "reference_wrapper for string" but does not give on "reference_wrapper for int"?
int main(){
int s= 43;
string str = "hello";
reference_wrapper<int> x{s};
reference_wrapper<string> y{str};
x.get() = 47;
y.get() = "there";
cout<<"printing original int "<<s<<"\n";
cout<<"printing original string "<<str<<"\n";
cout<<"printing reference_wrapper for int "<<x<<"\n";
cout<<"printing reference_wrapper for string "<<y<<"\n"; // gives error
int& refint = x;
string& refstr = y;
cout<<"printing reference for int "<<refint<<"\n";
cout<<"printing reference for string "<<refstr<<"\n";
}
operator<< for std::string is a function template, when being passed a reference_wrapper, the last template argument Allocator fails to be deduced; because implicit conversion won't be considered in template argument deduction.
Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.
As the workaround, you can call std::reference_wrapper<T>::get explicitly or perform explicit conversion.
On the other hand, operator<< for int is a non-template, then doesn't have such issue.