Copy local is important for deployment scenarios and tools. As a general rule you should use CopyLocal=True if the reference is not contained within the GAC. Copy Local essentially means I must manually deploy this DLL in order for my application to work. When it’s false it essentially means “I depend on another component which … Read more
Read this article: Appearing and Disappearing consts in C++ Type deduction for auto variables in C++0x is essentially the same as for template parameters. (As far as I know, the only difference between the two is that the type of auto variables may be deduced from initializer lists, while the types of template parameters may … Read more
You need to implement Add on &Vector rather than on Vector. impl<‘a, ‘b> Add<&’b Vector> for &’a Vector { type Output = Vector; fn add(self, other: &’b Vector) -> Vector { Vector { x: self.x + other.x, y: self.y + other.y, } } } In its definition, Add::add always takes self by value. But references … Read more
This is available from v1.9.0+. From NEWS: o Following this S.O. post, a function setDT is now implemented that takes a list (named and/or unnamed), data.frame (or data.table) as input and returns the same object as a data.table by reference (without any copy). See ?setDT examples for more. This is in accordance with data.table naming … Read more
There is no difference unless you write something weird like (v1 += v2) = v3; In the first case, the assignment will be to a temporary, and the overall effect will be v1 += v2. In the second case, the assignment will be to v1, so the overall effect will be v1 = v3. In … Read more
The reference itself is basically a pointer. 32 bits on a 32 bit OS, 64 bits on a 64 bit OS. The size of the object that’s referenced is more complicated.
This may seem counter-intuitive but I think the way to understand this is to realize that, in certain respects, references are treated syntactically like pointers. This seems logical for a pointer: int main() { boolalpha(cout); int const i = 1; cout << is_const<decltype(i)>::value << endl; int const* ri = &i; cout << is_const<decltype(ri)>::value << endl; … Read more
The reason that C++ does not allow you to rebind references is given in Stroustrup’s “Design and Evolution of C++” : It is not possible to change what a reference refers to after initialization. That is, once a C++ reference is initialized it cannot be made to refer to a different object later; it cannot … Read more
name_list2 = new List<string>(name_list1); This will clone the list. Edit: This solution only works for primitive types. For objects, see other responses below.
Also, check that the new solution’s projects run against a compatible framework to the project you’re trying to include. I was trying to include a reference to a 4.0 project in a 3.5 project.