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Quick Q: What's the difference between std::move and std::forward?

By Adrien Hamelin | Feb 27, 2018 07:50 PM | Tags: c++11 advanced

Quick A: One is used to forward parameters, one to move an object.

Recently on SO:

What's the difference between std::move and std::forward?

std::move takes an object and allows you to treat it as a temporary (an rvalue). Although it isn't a semantic requirement, typically a function accepting a reference to an rvalue will invalidate it. When you see std::move, it indicates that the value of the object should not be used afterwards, but you can still assign a new value and continue using it.

std::forward has a single use case: to cast a templated function parameter (inside the function) to the value category (lvalue or rvalue) the caller used to pass it. This allows rvalue arguments to be passed on as rvalues, and lvalues to be passed on as lvalues, a scheme called "perfect forwarding."

To illustrate:

void overloaded( int const &arg ) { std::cout << "by lvalue\n"; }
void overloaded( int && arg ) { std::cout << "by rvalue\n"; }

template< typename t >
/* "t &&" with "t" being template param is special, and  adjusts "t" to be
   (for example) "int &" or non-ref "int" so std::forward knows what to do. */
void forwarding( t && arg ) {
    std::cout << "via std::forward: ";
    overloaded( std::forward< t >( arg ) );
    std::cout << "via std::move: ";
    overloaded( std::move( arg ) ); // conceptually this would invalidate arg
    std::cout << "by simple passing: ";
    overloaded( arg );
}

int main() {
    std::cout << "initial caller passes rvalue:\n";
    forwarding( 5 );
    std::cout << "initial caller passes lvalue:\n";
    int x = 5;
    forwarding( x );
}

As Howard mentions, there are also similarities as both these functions simply cast to reference type. But outside these specific use cases (which cover 99.9% of the usefulness of rvalue reference casts), you should use static_cast directly and write a good explanation of what you're doing.

Quick Q: How can a class template store either reference or value?

By Adrien Hamelin | Feb 20, 2018 08:48 AM | Tags: advanced

Quick A: This happen by default.

Recently on SO:

How can a class template store either reference or value?

You already wrote it (minus the required template <typename T>). The deduction rules for a forwarding reference preserve value category as follows:

  1. If t is bound to an lvalue of type T2, then T = T2&.
  2. If t is bound to an rvalue of type T2, then T = T2.

It's those deduction rules that std::forward relies on to do its job. And why we need to pass the type to it as well.

The above means that you instantiate holder directly with T2 in the rvalue case. Giving you exactly what you want. A copy is made.

As a matter of fact, two copies are made. Once to create the constructor argument t, and the other copy is to initialize obj_m from it. But we can get rid of it with some clever use of type_traits:

template <class T>
class holder {
    T  obj_m;  // should be a reference if possible...
public:
    holder(std::add_rvalue_reference_t<T> t) :obj_m { std::forward<T>(t) } {}
};

template<typename T>
auto hold_this(T && t) { return holder<T>(std::forward<T>(t)); }

See it live. We use add_rvalue_reference_t to make t be of the correct reference type in each case. And "simulate" the argument deduction which would make obj_m { std::forward<T>(t) } resolve to initializing obj_m from the correct reference type.

I say "simulate" because it's important to understand the constructor argument for holder cannot be a forwarding reference because the constructor itself is not templated.

By the way, since you tagged c++17, we can also add a deduction guide to your example. If we define it as follows (with the feedback from T.C. incorporated):

template <class T>
class holder {
    T  obj_m;  // should be a reference if possible...
public:
    holder(T&& t) :obj_m { std::forward<T>(t) } {}
};

template<typename T>
holder(T&&) -> holder<T>;

Then this live example shows you can define variables as hold h1{t}; and hold h2{test()};, with the same deduced types as the function return values from before.

Quick Q: Most concise way to disable copy and move semantics

By Adrien Hamelin | Feb 20, 2018 08:40 AM | Tags: c++11 advanced

Quick A: Delete the move assignment.

Recently on SO:

Most concise way to disable copy and move semantics

According to this chart (by Howard Hinnant):

The most concise way is to =delete move assignment operator (or move constructor, but it can cause problems mentioned in comments).

Though, in my opinion the most readable way is to =delete both copy constructor and copy assignment operator.

The Most Vexing Parse: How to Spot It and Fix It Quickly--Jonathan Boccara

By Adrien Hamelin | Jan 31, 2018 07:22 PM | Tags: advanced

Did you have that problem?

The Most Vexing Parse: How to Spot It and Fix It Quickly

by Jonathan Boccara

From the article:

Everyone has their little defaults. You know, that little something that they do from time to time and that gets on your nerves, even though they’re otherwise nice people?

For C++, one of these little annoyances is the most vexing parse, well, as its name suggests...