Alex Rivera | Logout

Why override operator()?

Asked 2008-11-25T14:07:37.847
91

In the Boost Signals library, they are overloading the () operator.

Is this a convention in C++? For callbacks, etc.?

I have seen this in code of a co-worker (who happens to be a big Boost fan). Of all the Boost goodness out there, this has only led to confusion for me.

Any insight as to the reason for this overload?

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3 Answers

5

You may also look over the C++ faq's Matrix example. There are good uses for doing it but it of course depends on what you are trying to accomplish.

answered 2008-11-25T14:14:52.840
5

A functor is not a function, so you cannot overload it.
Your co-worker is correct though that the overloading of operator() is used to create "functors" - objects that can be called like functions. In combination with templates expecting "function-like" arguments this can be quite powerful because the distinction between an object and a function becomes blurred.

As other posters have said: functors have an advantage over plain functions in that they can have state. This state can be used over a single iteration (for example to calculate the sum of all elements in a container) or over multiple iterations (for example to find all elements in multiple containers satisfying particular criteria).

answered 2008-11-25T14:15:20.827
3

One strength I can see, however this can be discussed, is that the signature of operator() looks and behaves the same across different types. If we had a class Reporter which had a member method report(..), and then another class Writer, which had a member method write(..), we would have to write adapters if we would like to use both classes as perhaps a template component of some other system. All it would care about is to pass on strings or what have you. Without the use of operator() overloading or writing special type adapters, you couldn't do stuff like

T t;
t.write("Hello world");

because T has a requirement that there is a member function called write which accepts anything implicitly castable to const char* (or rather const char[]). The Reporter class in this example doesn't have that, so having T (a template parameter) being Reporter would fail to compile.

However, as far I can see this would work with different types

T t;
t("Hello world");

though, it still explicitly requires that the type T has such an operator defined, so we still have a requirement on T. Personally, I don't think it's too wierd with functors as they are commonly used but I would rather see other mechanisms for this behavior. In languages like C# you could just pass in a delegate. I am not too familiar with member function pointers in C++ but I could imagine you could achieve the same behaviour there aswell.

Other than syntatic sugar behaviour I don't really see the strengths of operator overloading to perform such tasks.

I am sure there are more knowingly people who have better reasons than I have but I thought I'd lay out my opinion for the rest of you to share.

answered 2008-11-25T14:37:48.020

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