C++ boost lexical_cast with template?

Question

I'm trying to build a class that stores program settings as a std::map. Since all the program settings are stored as strings I'd like an accessor method that can return the program setting casted to the relevant type. I'm new to templating in C++ and this is my first attempt:

class Settings
{
public:
    Settings(void);
    virtual ~Settings(void);

    enum SettingName {HomePageUrl, WindowWidth};

    template<class T>
    T Get(SettingName name)
    {
        return boost::lexical_cast<T>(settings_[name]);
    }

    template<class T>
    void Set(SettingName name, T value)
    {
        settings_[name] = boost::lexical_cast<CString>(value);
    }

private:
    std::map<SettingName, CString> settings_;

};

However, I'm getting a compiler errors:

...boost\boost_1_46_1\boost\lexical_cast.hpp(776): error C2678: binary '>>' :
no operator found which takes a left-hand operand of type
'std::basic_istream<_Elem,_Traits>' (or there is no acceptable conversion)

..settings.h(33) : see reference to function template instantiation
'Target boost::lexical_cast<CString,T>(const Source &)' being compiled

With boost the error output is very long and I'm not really sure what's wrong with it.

score 4 · Accepted Answer · answered Jun 17 '11 at 21:07

4

CString does not have any operator<< Consider using std::string

answered Jun 17 '11 at 21:07

Joel Falcou

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1

or provide such an operator for CString as lexical_cast internally use ostringstream to generate its output. – Joel Falcou Jun 17 '11 at 21:12

score 3 · Answer 2 · answered Jun 17 '11 at 21:18

binary '>>' : no operator found which takes a left-hand operand of type 'std::basic_istream<_Elem,_Traits>'

lexical_cast basically tries to write the object into a stream object.

you need << and >> operator defined to write to a stream in the class you're using for it to work. (depends if you're reading or writing)

score 3 · Answer 3 · answered Jun 17 '11 at 21:28

As shown in the documentation, boost::lexical_cast does its conversion based on the presence of several things. The source type must have an operator<< that takes a std::ostream (or std::wostream), and the destination type must have an operator>> that takes a std::istream (or std::wistream). The first parameter to these function is a non-const reference to the stream, and the second parameter is a reference to the type to send/construct.

In order to convert a setting name to a T, that T must have an operator>> that takes an input stream. Similarly, in order to convert to a CString, there must be an operator<< that takes an output stream.

C++ boost lexical_cast with template?

3 Answers3