c++0x: proper way to receive a lambda as parameter by reference

问题

What is the right way to define a function that receives a int->int lambda parameter by reference?

void f(std::function< int(int) >& lambda);

void f(auto& lambda);

I'm not sure the last form is even legal syntax.

Are there other ways to define a lambda parameter?

回答1:

You cannot have an auto parameter. You basically have two options:

Option #1: Use std::function as you have shown.

Option #2: Use a template parameter:

template<typename F>
void f(F &lambda) { /* ... */}

Option #2 may, in some cases, be more efficient, as it can avoid a potential heap allocation for the embedded lambda function object, but is only possible if f can be placed in a header as a template function. It may also increase compile times and I-cache footprint, as can any template. Note that it may have no effect as well, as if the lambda function object is small enough it may be represented inline in the std::function object.

回答2:

I would use template as:

template<typename Functor>
void f(Functor functor)
{
   cout << functor(10) << endl;
}

int g(int x)
{
    return x * x;
}
int main() 
{
    auto lambda = [] (int x) { cout << x * 50 << endl; return x * 100; };
    f(lambda); //pass lambda
    f(g);      //pass function 
}

Output:

500
1000
100

Demo : http://www.ideone.com/EayVq

回答3:

I know it's been 7 years, but here's a way nobody else mentioned:

void foo(void (*f)(int)){
    std::cout<<"foo"<<std::endl;
    f(1); // calls lambda which takes an int and returns void
}
int main(){
    foo([](int a){std::cout<<"lambda "<<a<<std::endl;});
}

Which outputs: