Run specific code when overloading C++ operators

Question

I have a class, let's call it Foo, which contains the 3 following methods (overloading the left-associative < binary operator):

... operator<(A a) { return *this; }
... operator<(B b) { return *this; }
... operator<(C c) { return *this; }

A, B, C classes are not related in any way(if that, is of any matter).

Now In my program I only have the 2 following cases:

A a = new A();
B b = new B();
C c = new C();

(First Case): new Foo() < a < b;

or

(Second Case): new Foo() < a < b < c;

Whenever I have the first case (which ends with b), I want to execute a function run() when I have read(I know) the b instance. So the idea is that I would have the following code in the Foo class:

... operator<(B b)
{
    run();
}

Now when I have the code of the first case, run() is being executed.

The problem is that when I have code like in the second case(which ends in c). I want to execute the run() function again but NOT until I know what c is. So If I have the previous piece of code run() will be called when doing < b which is not what I want as I don't know c yet. If I add run() in operator<(C c) I will call run() twice.

In a few words what I want to accomplish is when having the first case call run() at operator<(B b) and when I have the second case ONLY call run at operator<(C c).

Any ideas on how this can be solved(if it can)?

Overriding operators in non standard ways is frowned upon as it is non intuitive to use for the maintainer. Prefer to use methods with a good name that explains what is happening. — Martin York, Dec 20 '10 at 19:07
You are mixing the use of objects with pointers. The first step here is to understand the difference and then most of your problems will go away. — Martin York, Dec 20 '10 at 19:08
I'm with @Martin on this. You're violating [the basic rules of operator overloading](http://stackoverflow.com/questions/4421706/operator-overloading/4421708#4421708). Consider using functions instead. Oh, and what you're looking for is expression templates. — sbi, Dec 20 '10 at 19:39
(You should properly @address people in comment replies, so that your replies show up in their "Responses" tab.) No you do not _need_ to do that. I'm certain the way you want to do it will seem _much_ uglier to most seasoned C++ programmers than a solution that involves calling nicely named functions. Abusing operator overloading to make code "neater" basically _always_ fails. Hence the first two basic rules. — sbi, Dec 27 '10 at 22:20

score 2 · Answer 1 · answered Dec 20 '10 at 18:57

2

You could create separate proxy object return types, so when you define an operator for that proxy object, you know which types have gone before it.

answered Dec 20 '10 at 18:57

Puppy

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1

This is way beyond the OP ability level. There are many more fundamental problems that need to be solved first. – Martin York Dec 20 '10 at 19:09
2

My opinion is: If OP is so unstructured and wrong using instances and pointers then he/she is in the same way unstructured and wrong in it's overall software design. – mmmmmmmm Dec 20 '10 at 19:18

score 0 · Answer 2 · answered Dec 20 '10 at 19:58

You have to create a syntax tree before, and than evaluate it. So you are creating a DSL, domain specific language.

read this http://codeidol.com/cpp/cpp-template-metaprogramming/Domain-Specific-Embedded-Languages/-10.5.-Blitz-and-Expression-Templates/

or you can use the boost::proto

score 0 · Accepted Answer · edited Dec 20 '10 at 20:33

class PreFoo {
public:
    PreFoo() { ... }
    PreFoo & operator<<(A a) { ...; return *this; }
    PreFoo & operator<<(B b) { ...; return *this; }
    PreFoo & operator<<(C c) { ...; return *this; }

    int aa, bb, cc; // save your information
};

PreFoo MakeFoo() { return PreFoo(); }

class Foo {
public:
    Foo(const PreFoo & pre) { run(); } // implicit conversion

    void run() {} // your run!!
};


void g() {
  A a; B b; C c;
  // implicit conversion at the end
  Foo foo1 = MakeFoo() << a << b;
  Foo foo2 = PreFoo() << a << b << c;
}

Run specific code when overloading C++ operators

3 Answers3