How to pass by lambda in C++0x? - c++

Question

It seems the way to construct objects in C++0x avoiding copies/moves (particularly for large stack allocated objects) is "pass by lambda".
See the following code:
#include <iostream>
#define LAMBDA(x) [&] { return x; }
class A
{
public:
A() {};
A(const A&) { std::cout << "Copy "; }
A(A&&) { std::cout << "Move "; }
};
class B1
{
public:
B1(const A& a_) : a(a_) {}
B1(A&& a_) : a(std::move(a_)) {}
A a;
};
class B2
{
public:
B2(const A& a_) : a(a_) {}
B2(A&& a_) : a(std::move(a_)) {}
template <class LAMBDA_T>
B2(LAMBDA_T&& f, decltype(f())* dummy = 0) : a(f()) {}
A a;
};
int main()
{
A a;
std::cout << "B1 b11( a ): ";
B1 b11(a);
std::cout << std::endl;
std::cout << "B2 b12(LAMBDA(a)): ";
B2 b12(LAMBDA(a));
std::cout << std::endl;
std::cout << std::endl;
std::cout << "B1 b21( std::move(a) ): ";
B1 b21(std::move(a));
std::cout << std::endl;
std::cout << "B2 b22(LAMBDA(std::move(a))): ";
B2 b22(LAMBDA(std::move(a)));
std::cout << std::endl;
std::cout << std::endl;
std::cout << "B1 b31(( A() )): ";
B1 b31((A()));
std::cout << std::endl;
std::cout << "B2 b32((LAMBDA(A()))): ";
B2 b32((LAMBDA(A())));
std::cout << std::endl;
std::cout << std::endl;
}
Which outputs the following:
B1 b11( a ): Copy
B2 b12(LAMBDA(a)): Copy
B1 b21( std::move(a) ): Move
B2 b22(LAMBDA(std::move(a))): Move
B1 b31(( A() )): Move
B2 b32((LAMBDA(A()))):
Note the "pass by lambda" removes the move in the case where the parameter is a what I believe is called a "prvalue".
Note that it seems the "pass by lambda" approach only helps when the parameter is a "prvalue", but it doesn't seem to hurt in other cases.
Is there anyway to get functions to accept "pass by lambda" parameters in C++0x, that is nicer than the client having to wrap their parameters in lambda functions themselves? (other than defining a proxy macro that calls the function).

Answer 1

If you're okay with a templated constructor, you might as well use perfect forwarding instead of the obfuscation with lambdas.
class super_expensive_type {
public:
struct token_t {} static constexpr token = token_t {};
super_expensive_type(token_t);
}
constexpr super_expensive_type::token_t super_expensive_type::token;
class user {
public:
template<typename... Args>
explicit
user(Args&&... args)
: member { std::forward<Args>(args)... }
{}
private:
super_expensive_type member;
};
// ...
// only one construction here
user { super_expensive_type::token };
super_expensive_type moved_from = ...;
// one move
user { std::move(moved_from) };
super_expensive_type copied_from = ...;
// one copy
user { copied_from };
Using lambdas can't be better than this because the result from the expression in the lambda body has to be returned.

Answer 2

There's a fundamental problem with what you're doing. You cannot magic an object into existence. The variable must be:
Default constructed
Copy constructed
Move constructed
Constructed with a different constructor.
4 is off the table, since you only defined the first three. Your copy and move constructors both print things. Therefore, the only conclusion one can draw is that, if nothing is printed, the object is being default constructed. IE: filled with nothing.
In short, your Lambda-based transfer mechanism doesn't seem to be transferring anything at all.
After further analysis, I see what's happening. Your lambda isn't actually taking a value by reference; it's constructing a value. If you expand the macro, what you get is this:
B2 b32(([&] {return A()}));
It constructs a temporary; it doesn't actually take anything by reference. So I'm not sure how you can consider this "passing" anything. All you're doing is making a function that constructs an object. You could just as easily pass the arguments for B2::a's constructor to the constructor of B2 and have it use them to create the object, and it would give you the same effect.
You're not passing a value. You're making a function that will always create the exact same object. That's not very useful.