Initializing member class with non-default constructor

风流意气都作罢 提交于 2021-02-09 01:33:06

问题


I'm trying to make a gui which has a SimpleWindow class that contains a textPanel class:

class textPanel{
    private:
        std::string text_m;

    public:
        textPanel(std::string str):text_m(str){}
        ~textPanel();
};


class SimpleWindow{
    public:
        SimpleWindow();
        ~SimpleWindow();
        textPanel text_panel_m;
};

SimpleWindow::SimpleWindow():
        text_panel_m(std::string temp("default value"))
{
}

I want to be able to initialize the text_panel_m using a const char* that gets converted to a std::string without needing to make another constructor that takes a const char*. Should I just create another constructor with const char* as an argument anyway? If I do it this way is there a way to reduce the amount of redundant constructor code using c++0x?

With the approach above I'm having difficulties initializing the text_panel_m member variable. g++ gives me the following error:

simpleWindow.cpp:49: error: expected primary-expression before ‘temp’
simpleWindow.cpp: In member function ‘bool SimpleWindow::drawText(std::string)’:

How do I go about initializing the text_panel_m member variable without using the default constructor?


回答1:


You want an unnamed temporary value in the initializer list. One simple change will do it:

SimpleWindow::SimpleWindow():
         text_panel_m(std::string("default value"))



回答2:


You're almost there:

SimpleWindow::SimpleWindow():
        text_panel_m("default value")
{
}

Should do the trick, using std::string's implicit converting constructor from const char*.




回答3:


Change

text_panel_m(std::string temp("default value"))

to

text_panel_m(std::string("default value"))



回答4:


try str("string") and remove the std::string bit.

Alternatively you could have a default constructor on the textPanel class that calls your string constructor.



来源:https://stackoverflow.com/questions/5732820/initializing-member-class-with-non-default-constructor

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