Members vs method arguments access in C++

2019-02-15 08:06发布

Can I have a method which takes arguments that are denoted with the same names as the members of the holding class? I tried to use this:

    class Foo {
        public:
            int x, y;
            void set_values(int x, int y)
            {
                x = x;
                y = y;
            };
    };

... but it doesn't seem to work.

Is there any way of accessing the the instance the namespace of which I'm working in, similar to JavaScript's this or Python's self?

3条回答
▲ chillily
2楼-- · 2019-02-15 08:27

Yes, you should be able to write this using the "this" keyword (which is a pointer in C++):

class Foo {
    public:
        int x, y;
        void set_values(int x, int y)
        {
            this->x = x;
            this->y = y;
        }
}
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三岁会撩人
3楼-- · 2019-02-15 08:38

It's generally a good idea to avoid this kind of confusion by using a naming convention for member variables. For example, camelCaseWithUnderScore_ is quite common. That way you would end up with x_ = x;, which is still a bit funny to read out loud, but is fairly unambiguous on the screen.

If you absolutely need to have the variables and arguments called the same, then you can use the this pointer to be specific:

class Foo {
    public:
        int x, y;
        void set_values(int x, int y)
        {
            this->x = x;
            this->y = y;
        }
};

By the way, note the trailing semi-colon on the class definition -- that is needed to compile successfully.

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干净又极端
4楼-- · 2019-02-15 08:48

In C++ the current instance is referenced by the const pointer this.

class Foo {
    public:
        int x, y;
        void set_values(int x, int y)
        {
            this->x = x;
            this->y = y;
        };
};
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