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I want to pass a struct implementing operator() in to a function that accepts boost::function. This struct keeps track of the number of times it's been called.

struct CallCounter
{
    CallCounter() : count( 0 ) {}
    void operator()()
    {
        // do stuff
        cout << "count is at " << count << endl;
        ++count;
    }

    int count;
};

However, when I try to access count after passing it to another function, count is still at 0.

void callNTimes( int n, boost::function<void()> func )
{
    for ( int i = 0; i < n; ++i )
        func();
}

int main()
{
    CallCounter counter;
    callNTimes( 20, counter );

    cout << counter.count << endl; // prints 0
    return 0;
}

Even though while counter is being called, it's printing out the correct count. I understand that boost::function is making a copy of my struct counter. Is there a way to pass it by reference so that afterwards, count is the right number?

Brian Schlenker
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2 Answers2

3

The boost function stuff takes variables by value by default, much like normal functions. If you want the function to use a reference to your CallCounter, you have to tell it that explicitly:

callNTimes( 20, boost::ref(counter) );

See it here: http://coliru.stacked-crooked.com/a/5843b6dde685b569

There also exists a boost::cref if you want a constant reference, but no boost::rref (which would be nearly identical to a value, but single use. function is multi-use, so no moves). Additionally, as of C++11, all of this is also in the std namespace.

In addition, Jefffrey's suggestion is generally considered a better way to code this sort of thing, if possible. It can have a negative impact on compile times though.

Mooing Duck
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1

I see that you are using boost::function. It's actually not necessary here, you can just use templates and pass the functor by reference:

template<typename Functor>
void callNTimes( int n, Functor& func )
{
    for ( int i = 0; i < n; ++i )
        func();
}

Live demo

Shoe
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  • I think I'd recommend `Functor&&` in C++11, or a `const Functor&`/`Functor&` pair in C++03. – Mooing Duck Apr 17 '14 at 18:31
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    In this particular case between `const Functor&` and `Functor&` only the latter will work, and `Functor&&` doesn't really change much; but as a general idea, I definitely second the suggestion. – Shoe Apr 17 '14 at 19:30