CODEWITHSUNDEEP
c++functionfunction-pointers
Aryan
Aryan

Reputation: 638

Move reference (rvalue) to function

I was reading some documentation and saw this:

template<class Ret, class... Args>
struct is_function<Ret(Args...) &&> : std::true_type {};

Quoted from: http://en.cppreference.com/w/cpp/types/is_function

How can you have a rvalue reference to a function?

From what I understand, functions don't have storage lifetimes. Could someone explain this? I understand references and pointers but how can you "move" a function?

I wrote this code and it compiles and runs as it should:

#include <iostream>
using namespace std;

int foo(int num) {
    return num + 1;
}

int main() {

    int (*bar1)(int) = &foo;
    cout << bar1(1) << endl;

    int (&bar2)(int) = foo;
    cout << bar2(2) << endl;

    auto bar3 = std::move(bar2); // ????
    cout << bar3(3) << endl;
    cout << bar2(2) << endl;

    int (&&bar4)(int) = foo; // ????
    cout << bar4(4) << endl;

}

Lets just say if you could store functions as bytecode/opcodes in memory, and 'move' that around. Wouldn't the CPU prevent it from running?

EDIT: @NicolBolas corrected my misunderstanding, but here is an answer for my other 'question': rvalue reference to function

Upvotes: 0

Views: 169

Answers (1)

Nicol Bolas
Nicol Bolas

Reputation: 473352

How can you have a rvalue reference to a function?

That's not what that means.

The && at the end of Ret(Args...) && refers to the ability for a member function to have an rvalue this. So that specialization works for a function type that has Ret as the return value, Args as its arguments, and uses an rvalue this.

So it's not an "rvalue reference to a function". It's a function that takes an rvalue this.

Upvotes: 4

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