Move semantics and perfect forwarding difference - c++

I already got what move semantics is from this question:
What are move semantics?
But I still do not get what perfect forwarding is in relation to move semantics.
Can someone explain in simple english and with a simple example what perfect forwarding means?

Plain English-only attempt
The problem is probably too complex to be accurately described by plain English sentences, but one could think of perfect forwarding as a way to move temporary values passed to a function to another one as if the first function didn't exist at all, so without any unnecessary copies or assignments. C++11 allows you to do this by introducing some conversion rules between r-value (&&) and l-value (&) references to a type when you try to get a reference (either r-value or l-value) out of them.
R-value references are a feature of C++11 and they were designed to both address move semantics and perfect forwarding issues
This is the plain-English explanation but if you want to thoroughly understand the problem, I'd suggest reading the following:
The problem:
We want some temporary values passed to a function F to be passed to another one E without any copy or assignment.
Attempts to solve this issue
If you try to pass it by reference like
template<typename T> void F(T& a) { E(a); }
you will not be able to use temporaries (they're not l-values)
F(1, 2, 3); // Won't work
Declaring a reference as const prolongs the lifetime of a temporary on the stack (this was historically done to avoid a common dangling reference error) so the following works
template<typename T> void E(const T& a) {}
template<typename T> void F(const T& a) {
E(a);
}
but the downside is that you'll have to modify the signature of the function(s) to conform to this solution
If we're interested in the signature of E (it should conform to something) but not in F's one, we might get away with
template<typename T> void E(T& a) {}
template<typename T> void F(const T& a) {
E(const_cast<T&>(a));
}
but in case this gets called with a real const and gets un-constant'ed, that would trigger undefined behavior
An unmaintainable solution could be to define all the variants you need
template<typename T> void E(T& a) {}
template<typename T> void F(T& a) { E(a); }
template<typename T> void F(const T& a) { E(const_cast<T&>(a)); }
but as the number of parameters grow, the number of combinations grows as well: this is likely to become unmaintainable
The solution in C++11
C++11 defines some rules that state
"[given] a type TR that is a reference to a type T, an attempt to
create the type “lvalue reference to cv TR” creates the type “lvalue
reference to T”, while an attempt to create the type “rvalue reference
to cv TR” creates the type TR."
in human-form (TR = a reference to type T, R = reference):
TR R
T& & -> T& // an lvalue reference to cv TR (becomes)-> lvalue reference to T
T& && -> T& // an rvalue reference to cv TR (becomes)-> TR (lvalue reference to T)
T&& & -> T& // an lvalue reference to cv TR (becomes)-> lvalue reference to T
T&& && -> T&& // an rvalue reference to cv TR (becomes)-> TR (rvalue reference to T)
The important takeaway here is that now you can keep track of the type the function received: you can receive an l-value and pass the same l-value to E or you can receive an r-value and pass the same r-value (after converting it since an l-value reference to whatever type reference becomes an l-value reference) to E:
template<typename T> void E(T&& a) {}
template<typename T> void F(T&& a) { E(static_cast<T&&>(a)); }
A synctactic sugar for
static_cast<T&&>(a)
is
std::forward<T>(a); // is the same as static_cast<T&&>(a);
so the final code that solves the problem and makes your life easier is
template<typename T> void E(T&& a) {}
template<typename T> void F(T&& a) { E(std::forward<T>(a)); }
Live example
References: Herb Sutter's blog and some other sources which unfortunately I can't find anymore. If anyone has a clue about those please write them in the comments below and I'll update the post. Thanks.

Dealing with r-value references and reference collapsing can be more complex than it initially appears.
Perfect forwarding
Perfect forwarding is there to ensure that the argument provided to a function is forwarded (passed) to another function with the same value category (basically r-value vs l-value) as originally provided.
It is typically used with template functions where reference collapsing may have taken place.
It can also be used within the same function.
Scott Meyers gives the following pseudo code in his Going Native 2013 presentation to explain the workings of std::forward (at approximately the 20 minute mark);
template <typename T>
T&& forward(T&& param) { // T&& here is formulated to disallow type deduction
if (is_lvalue_reference<T>::value) {
return param; // return type T&& collapses to T& in this case
}
else {
return move(param);
}
}
Example
An example from the site above, an archetypical example is that of make_unique
template<class T, class... U>
std::unique_ptr<T> make_unique(U&&... u)
{
return std::unique_ptr<T>(new T(std::forward<U>(u)...));
}
In the example, the arguments for the unique_ptr are provided to it through the make_unique as if they had been provided directly to unique_ptr, i.e. the reference, l-value and r-value nature of the arguments are maintained.
A more concrete example;
#include <iostream>
#include <utility>
#include <memory>
struct A {
// implementation excluded
};
struct B {
B(A &) // ctor 1
{
std::cout << "ctor 1" << std::endl;
}
B(A&&) // ctor 2
{
std::cout << "ctor 2" << std::endl;
}
};
int main()
{
A a;
auto b1 = std::make_unique<B>(a); // ctor 1 is used
auto b2 = std::make_unique<B>(A()); // ctor 2 is used
}
In Brief
Perfect forwarding depends on a handful of fundamental language constructs new to C++11 that form the bases for much of what we now see in generic programming:
Reference collapsing
Rvalue references
Move semantics
The use of std::forward is currently intended in the formulaic std::forward<T>, understanding how std::forward works helps understand why this is such, and also aids in identifying non-idiomatic or incorrect use of rvalues, reference collapsing and ilk.
Thomas Becker provides a nice, but dense write up on the perfect forwarding problem and solution.

Related

'&&' in function parameter pack

I have seen multiple instances of code where function parameter pack is declared using the && notation, as shown below, but I cannot see any advantage to using this notation.
template<typename... Args>
void Function(Args... args)
{
}
template<typename... Args>
void Function(Args&&... args)
{
}
My first thought was that the && form will be used exclusively for r-value objects, but this test proved that wrong:
struct Object
{
// Added bodies so I see what is being called via a step-into
Object() {}
Object(const Object&) {}
Object(Object&&) noexcept {}
Object& operator=(const Object&) { return *this; }
Object& operator=(Object&&) noexcept { return *this; }
};
Object GetObject() { Object o; return o; }
Object obj;
Function(GetObject());
Function(GetObject());
Here, VS 2017 complains that both forms of the function are viable candidates for the call.
Can someone explain what the difference is between these two, and what advantages one may have over the other please?
They are forwarding references in the parameter pack form. As for template parameter deduction, they can match any arguments, but the template parameter will be deduced differently comparing to the ordinary template parameter.
The major advantage of forwarding reference is that the lvalue/rvalue information will be preserved if used with std::forward. Thus they are used to "forward" something.
For example,
void real_foo(A const &a);
void real_foo(A &&a);
template<class... Args>
void foo_proxy_ordinary(Args... args) { real_foo(args...); }
template<class... Args>
void foo_proxy_perfect(Args&&... args) { real_foo(std::forward<Args>(args)...); }
The ordinary version will always call real_foo(A const &) version, because inside foo_proxy, args are always lvalue.
However, the perfect version will select real_foo(A&&) if the arguments passed in are indeed rvalues.
Combining forwarding reference with parameter pack, one can write easily generic proxy functions without performance loss in terms of lvalue/rvalue.
T&& when used in the context of
template<typename T>
void f(T&& t);
is called a forwarding reference sometimes also called a universal reference.
Main advantage of a forwarding reference is that combined with std::forward it enables achieving a so-called perfect forwarding: function template passing its arguments to another function as they are (lvalue as lvalue, rvalue as rvalue).
Now it is possible to create higher-order functions that take other functions as arguments or return them, or superior function-wrappers (e.g., std::make_shared), and do other cool things.
Here is some material that explains it much better and in more detail than I possibly can:
Perfect forwarding and universal references in C++
Rvalue References and Perfect Forwarding in C++0x
Forwarding references proposal
SO: Advantages of using forward
SO: Perfect forwarding - what's it all about?
Can someone explain what the difference is between these two, and what advantages one may have over the other please?
The difference is same for parameter packs as it is for individual parameters. Args declares an "object parameter" (pass by value) and Args&& declares a reference parameter (pass by reference).
Passing by reference allows one to avoid copying the argument when that is unnecessary. It also allows modifying the referred argument if the reference is non-const, which includes the possibility of moving from that object.
Passing by value makes it clear to the caller that the passed object will neither be modified, nor be referred to as a result of calling the function.
My first thought was that the && form will be used exclusively for r-value objects
As your test demonstrates, that is indeed an incorrect assumption. When Args is a deduced type i.e. auto or a template argument, Args&& can indeed be either an l-value reference or an r-value reference. Which one it is depends on what Args is deduced to be. This demonstrates the reference collapsing rules concisely:
typedef int& lref;
typedef int&& rref;
int n;
lref& r1 = n; // type of r1 is int&
lref&& r2 = n; // type of r2 is int& note this case in particular
rref& r3 = n; // type of r3 is int&
rref&& r4 = 1; // type of r4 is int&&
Using such reference allows forwarding i.e. re-binding into a new lvalue reference (when possible) or moving from the object (when possible) or copying (when neither of the previous is possible).
Because of this, Args&& is called a forwarding reference (or a universal reference) when Args is a deduced type.

How is the move constructor of member variable invoked without using std::forward?

An example here for std::forward,
// forward example
#include <utility> // std::forward
#include <iostream> // std::cout
// function with lvalue and rvalue reference overloads:
void overloaded (const int& x) {std::cout << "[lvalue]";}
void overloaded (int&& x) {std::cout << "[rvalue]";}
// function template taking rvalue reference to deduced type:
template <class T> void fn (T&& x) {
overloaded (x); // always an lvalue
overloaded (std::forward<T>(x)); // rvalue if argument is rvalue
}
int main () {
int a;
std::cout << "calling fn with lvalue: ";
fn (a);
std::cout << '\n';
std::cout << "calling fn with rvalue: ";
fn (0);
std::cout << '\n';
return 0;
}
Output:
calling fn with lvalue: [lvalue][lvalue]
calling fn with rvalue: [lvalue][rvalue]
mentions that
the fact that all named values (such as function parameters) always
evaluate as lvalues (even those declared as rvalue references)
Whereas, the typical move constructor looks like
ClassName(ClassName&& other)
: _data(other._data)
{
}
which looks like _data(other._data) should invoke the move constructor of _data's class. But, how is it possible without using std::forward? In other words, shouldn't it be
ClassName(ClassName&& other)
: _data(std::forward(other._data))
{
}
?
Because, as pointed out in std:forward case,
all then named values should evaluate as lvalue
I more and more like C++ because of the depth of issue like this and the fact that the language is bold enough to provide such features :) Thank you!
A typical move constructor looks like this (assuming it is explicitly implemented: you might want to prefer = default):
ClassName::ClassName(ClassName&& other)
: _data(std::move(other._data)) {
}
Without the std::move() the member is copied: since it has a name other is an lvalue. The object the reference is bound to is an rvalue or an object considered as such, however.
std::forward<T>(obj) is always used with an explicit template argument. In practice the type is that deduced for a forwarding reference. These look remarkably like rvalue references but are something entirely different! In particular, a forwarding reference may refer to an lvalue.
You may be interested in my Two Daemons article which describes the difference in detail.
std::forward should be used with a forwarding reference.
std::move should be used with an rvalue reference.
There is nothing particular about constructors. The rules apply the same to any function, member function or constructor.
The most important thing is to realize when you have a forwarding reference and when you have an rvalue reference. They look similar but are not.
A forwarding reference is always in the form:
T&& ref
for T some deduced type.
For instance, this is a forwarding reference:
template <class T>
auto foo(T&& ref) -> void;
All these are rvalue references:
auto foo(int&& ref) -> void; // int not deduced
template <class T>
auto foo(const T&& ref); // not in form `T&&` (note the const)
template <class T>
auto foo(std::vector<T>&& ref) -> void; // not in form `T&&`
template <class T>
struct X {
auto foo(T&& ref) -> T; // T not deduced. (It was deduced at class level)
};
For more please check this excellent in-depth article by Scott Meyers with the note that when the article was written the term "universal reference" was used (actually introduced by Scott himself). Now it is agreed that "forwarding reference" better describes it's purpose and usage.
So your example should be:
ClassName(ClassName&& other)
: _data(std::move(other._data))
{
}
as other is an rvalue reference because ClassName is not a deduced type.
This Ideone example should make things pretty clear for you. If not, keep reading.
The following constructor accepts Rvalues only. However, since the argument "other" got a name it lost its "rvalueness" and now is a Lvalue. To cast it back to Rvalue, you have to use std::move. There's no reason to use std::forward here because this constructor does not accept Lvalues. If you try to call it with a Lvalue, you will get compile error.
ClassName(ClassName&& other)
: _data(std::move(other._data))
{
// If you don't use move, you could have:
// cout << other._data;
// And you will notice "other" has not been moved.
}
The following constructor accepts both Lvalues and Rvalues. Scott Meyers called it "Universal Rerefences", but now it's called "Forwarding References". That's why, here, it's a must to use std::forward so that if other was an Rvalue, _data constructor will get called with an Rvalue. If other was an Lvalue, _data will be constructed with an Lvalue. That's why it's called perfect-forwarding.
template<typename T>
ClassName(T&& other)
: _data(std::forward<decltype(_data)>(other._data))
{
}
I've tried to use your constructors as an example so you could understand, but this is not specific to constructors. This applies to functions as well.
With the first example tho, since your first constructor only accepts Rvalues, you could perfectly use std::forward instead, and both would do the same thing. But it's best not to do it, because people may think that your constructor accepts a forwarding reference, when it actually doesn't.

Why use a perfectly forwarded value (a functor)?

C++11 (and C++14) introduces additional language constructs and improvements that target generic programming. These include features such as;
R-value references
Reference collapsing
Perfect forwarding
Move semantics, variadic templates and more
I was browsing an earlier draft of the C++14 specification (now with updated text) and the code in an example in §20.5.1, Compile-time integer sequences, that I found interesting and peculiar.
template<class F, class Tuple, std::size_t... I>
decltype(auto) apply_impl(F&& f, Tuple&& t, index_sequence<I...>) {
return std::forward<F>(f)(std::get<I>(std::forward<Tuple>(t))...);
}
template<class F, class Tuple>
decltype(auto) apply(F&& f, Tuple&& t) {
using Indices = make_index_sequence<std::tuple_size<Tuple>::value>;
return apply_impl(std::forward<F>(f), std::forward<Tuple>(t), Indices());
}
Online here [intseq.general]/2.
Question
Why was the function f in apply_impl being forwarded, i.e. why std::forward<F>(f)(std::get...?
Why not just apply the function as f(std::get...?
In Brief...
The TL;DR, you want to preserve the value category (r-value/l-value nature) of the functor because this can affect the overload resolution, in particular the ref-qualified members.
Function definition reduction
To focus on the issue of the function being forwarded, I've reduced the sample (and made it compile with a C++11 compiler) to;
template<class F, class... Args>
auto apply_impl(F&& func, Args&&... args) -> decltype(std::forward<F>(func)(std::forward<Args>(args)...)) {
return std::forward<F>(func)(std::forward<Args>(args)...);
}
And we create a second form, where we replace the std::forward(func) with just func;
template<class F, class... Args>
auto apply_impl_2(F&& func, Args&&... args) -> decltype(func(std::forward<Args>(args)...)) {
return func(std::forward<Args>(args)...);
}
Sample evaluation
Evaluating some empirical evidence of how this behaves (with conforming compilers) is a neat starting point for evaluating why the code example was written as such. Hence, in addition we will define a general functor;
struct Functor1 {
int operator()(int id) const
{
std::cout << "Functor1 ... " << id << std::endl;
return id;
}
};
Initial sample
Run some sample code;
int main()
{
Functor1 func1;
apply_impl_2(func1, 1);
apply_impl_2(Functor1(), 2);
apply_impl(func1, 3);
apply_impl(Functor1(), 4);
}
And the output is as expected, independent of whether an r-value is used Functor1() or an l-value func when making the call to apply_impl and apply_impl_2 the overloaded call operator is called. It is called for both r-values and l-values. Under C++03, this was all you got, you could not overload member methods based on the "r-value-ness" or "l-value-ness" of the object.
Functor1 ... 1
Functor1 ... 2
Functor1 ... 3
Functor1 ... 4
Ref-qualified samples
We now need to overload that call operator to stretch this a little further...
struct Functor2 {
int operator()(int id) const &
{
std::cout << "Functor2 &... " << id << std::endl;
return id;
}
int operator()(int id) &&
{
std::cout << "Functor2 &&... " << id << std::endl;
return id;
}
};
We run another sample set;
int main()
{
Functor2 func2;
apply_impl_2(func2, 5);
apply_impl_2(Functor2(), 6);
apply_impl(func2, 7);
apply_impl(Functor2(), 8);
}
And the output is;
Functor2 &... 5
Functor2 &... 6
Functor2 &... 7
Functor2 &&... 8
Discussion
In the case of apply_impl_2 (id 5 and 6), the output is not as may have been initially been expected. In both cases, the l-value qualified operator() is called (the r-value is not called at all). It may have been expected that since Functor2(), an r-value, is used to call apply_impl_2 the r-value qualified operator() would have been called. The func, as a named parameter to apply_impl_2, is an r-value reference, but since it is named, it is itself an l-value. Hence the l-value qualified operator()(int) const& is called in both the case of the l-value func2 being the argument and the r-value Functor2() being used as the argument.
In the case of apply_impl (id 7 and 8) the std::forward<F>(func) maintains or preserves the r-value/l-value nature of the argument provided for func. Hence the l-value qualified operator()(int) const& is called with the l-value func2 used as the argument and the r-value qualified operator()(int)&& when the r-value Functor2() is used as the argument. This behaviour is what would have been expected.
Conclusions
The use of std::forward, via perfect forwarding, ensures that we preserve the r-value/l-value nature of the original argument for func. It preserves their value category.
It is required, std::forward can and should be used for more than just forwarding arguments to functions, but also when the use of an argument is required where the r-value/l-value nature must be preserved. Note; there are situations where the r-value/l-value cannot or should not be preserved, in these situations std::forward should not be used (see the converse below).
There are many examples popping up that inadvertently lose the r-value/l-value nature of the arguments via a seemingly innocent use of an r-value reference.
It has always been hard to write well defined and sound generic code. With the introduction of r-value references, and reference collapsing in particular, it has become possible to write better generic code, more concisely, but we need to be ever more aware of what the original nature of the arguments provided are and make sure that they are maintained when we use them in the generic code we write.
Full sample code can be found here
Corollary and converse
A corollary of the question would be; given reference collapsing in a templated function, how is the r-value/l-value nature of the argument maintained? The answer - use std::forward<T>(t).
Converse; does std::forward solve all your "universal reference" problems? No it doesn't, there are cases where it should not be used, such as forwarding the value more than once.
Brief background to perfect forwarding
Perfect forwarding may be unfamiliar to some, so what is perfect forwarding?
In brief, perfect forwarding is there to ensure that the argument provided to a function is forwarded (passed) to another function with the same value category (basically r-value vs. l-value) as originally provided. It is typically used with template functions where reference collapsing may have taken place.
Scott Meyers gives the following pseudo code in his Going Native 2013 presentation to explain the workings of std::forward (at approximately the 20 minute mark);
template <typename T>
T&& forward(T&& param) { // T&& here is formulated to disallow type deduction
if (is_lvalue_reference<T>::value) {
return param; // return type T&& collapses to T& in this case
}
else {
return move(param);
}
}
Perfect forwarding depends on a handful of fundamental language constructs new to C++11 that form the bases for much of what we now see in generic programming:
Reference collapsing
Rvalue references
Move semantics
The use of std::forward is currently intended in the formulaic std::forward<T>, understanding how std::forward works helps understand why this is such, and also aids in identifying non-idiomatic or incorrect use of rvalues, reference collapsing and ilk.
Thomas Becker provides a nice, but dense write up on the perfect forwarding problem and solution.
What are ref-qualifiers?
The ref-qualifiers (lvalue ref-qualifier & and rvalue ref-qualifier &&) are similar to the cv-qualifiers in that they (the ref-qualified members) are used during overload resolution to determine which method to call. They behave as you would expect them to; the & applies to lvalues and && to rvalues. Note: Unlike cv-qualification, *this remains an l-value expression.
Here is a practical example.
struct concat {
std::vector<int> state;
std::vector<int> const& operator()(int x)&{
state.push_back(x);
return state;
}
std::vector<int> operator()(int x)&&{
state.push_back(x);
return std::move(state);
}
std::vector<int> const& operator()()&{ return state; }
std::vector<int> operator()()&&{ return std::move(state); }
};
This function object takes an x, and concatenates it to an internal std::vector. It then returns that std::vector.
If evaluated in an rvalue context it moves to a temporary, otherwise it returns a const& to the internal vector.
Now we call apply:
auto result = apply( concat{}, std::make_tuple(2) );
because we carefully forwarded our function object, only 1 std::vector buffer is allocated. It is simply moved out to result.
Without the careful forwarding, we end up creating an internal std::vector, and we copy it to result, then discard the internal std::vector.
Because the operator()&& knows that the function object should be treated as a rvalue about to be destroyed, it can rip the guts out of the function object while doing its operation. The operator()& cannot do this.
Careful use of perfect forwarding of function objects enables this optimization.
Note, however, that there is very little use of this technique "in the wild" at this point. Rvalue qualified overloading is obscure, and doing so to operator() moreso.
I could easily see future versions of C++ automatically using the rvalue state of a lambda to implicitly move its captured-by-value data in certain contexts, however.

Can I typically/always use std::forward instead of std::move?

I've been watching Scott Meyers' talk on Universal References from the C++ and Beyond 2012 conference, and everything makes sense so far. However, an audience member asks a question at around 50 minutes in that I was also wondering about. Meyers says that he does not care about the answer because it is non-idiomatic and would silly his mind, but I'm still interested.
The code presented is as follows:
// Typical function bodies with overloading:
void doWork(const Widget& param) // copy
{
// ops and exprs using param
}
void doWork(Widget&& param) // move
{
// ops and exprs using std::move(param)
}
// Typical function implementations with universal reference:
template <typename T>
void doWork(T&& param) // forward => copy and move
{
// ops and exprs using std::forward<T>(param)
}
The point being that when we take an rvalue reference, we know we have an rvalue, so we should std::move it to preserve the fact that it's an rvalue. When we take a universal reference (T&&, where T is a deduced type), we want std::forward to preserve the fact that it may have been an lvalue or an rvalue.
So the question is: since std::forward preserves whether the value passed into the function was either an lvalue or an rvalue, and std::move simply casts its argument to an rvalue, could we just use std::forward everywhere? Would std::forward behave like std::move in all cases where we would use std::move, or are there some important differences in behaviour that are missed out by Meyers' generalisation?
I'm not suggesting that anybody should do it because, as Meyers correctly says, it's completely non-idiomatic, but is the following also a valid use of std::move:
void doWork(Widget&& param) // move
{
// ops and exprs using std::forward<Widget>(param)
}
The two are very different and complementary tools.
std::move deduces the argument and unconditionally creates an rvalue expression. This makes sense to apply to an actual object or variable.
std::forward takes a mandatory template argument (you must specify this!) and magically creates an lvalue or an rvalue expression depending on what the type was (by virtue of adding && and the collapsing rules). This only makes sense to apply to a deduced, templated function argument.
Maybe the following examples illustrate this a bit better:
#include <utility>
#include <memory>
#include <vector>
#include "foo.hpp"
std::vector<std::unique_ptr<Foo>> v;
template <typename T, typename ...Args>
std::unique_ptr<T> make_unique(Args &&... args)
{
return std::unique_ptr<T>(new T(std::forward<Args>(args)...)); // #1
}
int main()
{
{
std::unique_ptr<Foo> p(new Foo('a', true, Bar(1,2,3)));
v.push_back(std::move(p)); // #2
}
{
v.push_back(make_unique<Foo>('b', false, Bar(5,6,7))); // #3
}
{
Bar b(4,5,6);
char c = 'x';
v.push_back(make_unique<Foo>(c, b.ready(), b)); // #4
}
}
In situation #2, we have an existing, concrete object p, and we want to move from it, unconditionally. Only std::move makes sense. There's nothing to "forward" here. We have a named variable and we want to move from it.
On the other hand, situation #1 accepts a list of any sort of arguments, and each argument needs to be forwarded as the same value category as it was in the original call. For example, in #3 the arguments are temporary expressions, and thus they will be forwarded as rvalues. But we could also have mixed in named objects in the constructor call, as in situation #4, and then we need forwarding as lvalues.
Yes, if param is a Widget&&, then the following three expressions are equivalent (assuming that Widget is not a reference type):
std::move(param)
std::forward<Widget>(param)
static_cast<Widget&&>(param)
In general (when Widget may be a reference), std::move(param) is equivalent to both of the following expressions:
std::forward<std::remove_reference<Widget>::type>(param)
static_cast<std::remove_reference<Widget>::type&&>(param)
Note how much nicer std::move is for moving stuff. The point of std::forward is that it mixes well with template type deduction rules:
template<typename T>
void foo(T&& t) {
std::forward<T>(t);
std::move(t);
}
int main() {
int a{};
int const b{};
//Deduced T Signature Result of `forward<T>` Result of `move`
foo(a); //int& foo(int&) lvalue int xvalue int
foo(b); //int const& foo(int const&) lvalue int const xvalue int const
foo(int{});//int foo(int&&) xvalue int xvalue int
}

std::forward vs std::move while binding lvalue to rvalue reference

is there a difference between move and forward here:
void test(int && val)
{
val=4;
}
void main()
{
int nb;
test(std::forward<int>(nb));
test(std::move(nb));
std::cin.ignore();
}
In your specific case, no, there isn't any difference.
Detailed answer:
Under the hood, std::move(t) does static_cast<typename std::remove_reference<T>::type&&>(t), where T is type of t (see §20.2.3/6). In your case, it resolves to static_cast<int&&>(nb).
forward is a little bit tricky, because it is tailored for use in templates (to allow perfect forwarding) and not as a tool to cast lvalue to rvalue reference.
Standard library provides two overloads (one for lvalue references and the second for rvalue ones, see §20.2.3/2):
template <class T> T&& forward(typename remove_reference<T>::type& t) noexcept;
template <class T> T&& forward(typename remove_reference<T>::type&& t) noexcept;
Substituting int, we get:
int&& forward(int& t) noexcept;
int&& forward(int&& t) noexcept;
And since nb is lvalue, the first version is chosen. According to standard draft, the only effect of forward is static_cast<T&&>(t). With T being int, we get static_cast<int&&>(nb), i.e. - we get two exactly same casts.
Now, if you want to cast lvalue to rvalue (to allow moving), please use only std::move, which is the idiomatic way to do this conversion. std::forward is not intended to be used this way.
No difference.