Variadic template operator

问题

I'm trying to change some of my functions foo() into operator<<(), simply for the sake of getting some "half C/half C++" code to look more like C++. Happens, though, I've got stuck at the following transformation step:

template <class... T>
inline const size_t foo(const T&... data) {
    return sizeof...(T);
}
struct bar {
    template <class... T>
    inline const size_t operator<<(const T&... data) {
        return sizeof...(T);
    }
};
int main(int argc, char *argv[]) {
    bar a;
    std::cout << ">>> length " << foo(1, 2, 3) << std::endl;
    std::cout << ">>> length " << (a << 1 << 2) << std::endl;
    std::cout << ">>> length " << (a << 1 << 2 << 3) << std::endl;
    std::cout << ">>> length " << (a << 1 << 2 << 3 << 4) << std::endl;
}

From the output:

$ ./a.out 
>>> length 3
>>> length 4
>>> length 32
>>> length 512

I conclude the first computation is performed upon a << 1, and the subsequent values are shifted accordingly. Yet, I'm failing to see how I could rewrite foo(), as to provide a operator<<() interface to users of struct bar -- of course, without changing foo() semantics.

In case there's no way to pass class T... as parameter to operator<<(), the function would be naturally less efficient than foo(), as it would be called many times. Is there any reasonable C++ construct for this, or sticking to foo() is the only/best option here?

Context:

These foo() functions are senders/receivers to network communication. I thought that would be better to provide a more "C++" interface, with a sender/receiver stream, writable/readable using << and >> operators -- other than using regular functions foo(...).

回答1:

The language is doing what you are asking it to do.

With associativity, the following are equivalent (<< and >> are left-to-right associative):

a << 1 << 2
(a << 1) << 2

The call a << 1 calls your user-defined operator, which in turn returns a size_t. That's why the types for next call are the following: size_t << int (which is a simple bitwise shift).

You need to use expression templates. The idea is the following (live example here):

template<typename... args>
struct stream_op
{
};

template<typename... A, typename B>
stream_op<A..., B> operator<<(stream_op<A...> a, B b)
{
    // Do stuff
}

So, the following occur (with a as a stream_op<>):

a << 1 << 2
------
  |
  v
-------------------------------------------
stream_op<int> operator<<(stream_op<>, int) << 2
--------------                                ---
     |                                         |
     |             +---------------------------+
     v             v
--------------    ---
stream_op<int> << int
--------------    ---
       |           |
       |           +---------------------------+
       +----------------------------+          |
                                    v          v
                              --------------  ---
stream_op<int,int> operator<<(stream_op<int>, int)
------------------
        |
        v
------------------
stream_op<int,int> // <- Expected result

Then you just have to put a method to convert stream_op to int (or to whatever you want).

A note on performances: with these expression templates, a part of the data is encoded in the type, so normally it should be as fast as a direct call to foo(...).

来源：https://stackoverflow.com/questions/17868718/variadic-template-operator

标签

c++

operator-overloading

variadic-templates

Variadic template operator<<

问题

回答1: