我如何建立这种类型的类接口

Question

I am interested in trying something where I create a custom type and then access to its members using dot semantics. For example:

 Class A{ //simplified, omitting constructors and other methods
   private:
   float numbers[3];
   public:
   float x(){ return numbers[0]; }
   float y(){ return numbers[1]; }
   float z(){ return numbers[2]; }
  }

So I can do something like this:

  A a;
  //do stuff to populate `numbers`

  float x=a.x;

But I would also like to make the elements in numbers lvalues so I can do something like this:

  A a;
  a.y=5; //assigns 5 to numbers[1]

How can I do this setting method?

Answer 1

First. You made functions x, y and z but assigning them to float. This wouldn't work. Second. Change these functions to return referencies:

class A{ //simplified, omitting constructors and other methods
   private:
   float numbers[3];
   public:
   float & x(){ return numbers[0]; }
   float & y(){ return numbers[1]; }
   float & z(){ return numbers[2]; }
};
...
A point;
float x = point.x();
point.x() = 42.0f;

There's another way: declare referencies as a members of class and initialize them in c-tor:

class A{ //simplified, omitting constructors and other methods
   private:
   float numbers[3];
   public:
   float & x;
   float & y;
   float & z;
   A() : x( numbers[ 0 ] ), y( numbers[ 1 ] ), z( numbers[ 2 ] ) {}
};
...
A point;
float x = point.x;
point.x = 42.0f;

PS Pay an attention on comment, that gave @MikeSeymour

Answer 2

You can return a reference to allow assignment:

float & x(){ return numbers[0]; }
      ^

// usage
A a;
a.x() = 42;

You should also have a const overload, to allow read-only access to a const object:

float x() const {return numbers[0];}
          ^^^^^

// usage
A const a = something();
float x = a.x();

Answer 3

Not unless you actually have public variables named x, y and z.

Or you can return a reference and then do ay() = 5