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

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
}