I have a Vector:
struct vector { double coord[3]; };
I can do:
Vector v = q; // q is a qvm::vec
But how to do : q = v;?
The only way I found is to do qvm::assign (q, v);. Otherwise the compiler tells me:
no operator found which takes a right-hand operand of type Vector*.
I followed the documentation to define a trait (docs). It is probably what allows the assignment v = q to work.
How to do q = v? (if at all possible)?
The assign call is the approach to take (although you can use convert_to<T> as well).
The reason is implicitly given as part of the FAQ:
Exactly the same reasoning goes for operator= which can, also, only be declared inside the type itself.
You could extend the interface by inheriting, but that removes the POD-ness, which kind of defeats the purpose of using QVM (which is to avoid overhead).
Related
std::move() is stealing the string value whereas not an int, please help me.
int main()
{
int i = 50;
string str = "Mahesh";
int j = std::move(i);
string name = std::move(str);
std::cout <<"i: "<<i<<" J: "<<j <<std::endl;
std::cout <<"str: "<<str<<" name: "<<name <<std::endl;
return 0;
}
Output
i: 50 J: 50
str: name: Mahesh
std::move is a cast to an rvalue reference. This can change overload resolution, particularly with regard to constructors.
int is a fundamental type, it doesn't have any constructors. The definition for int initialisation does not care about whether the expression is const, volatile, lvalue or rvalue. Thus the behaviour is a copy.
One reason this is the case is that there is no benefit to a (destructive) move. Another reason is that there is no such thing as an "empty" int, in the sense that there are "empty" std::strings, and "empty" std::unique_ptrs
std::move() itself doesn't actually do any moving. It is simply used to indicate that an object may be moved from. The actual moving must be implemented for the respective types by a move constructor/move assignment operator.
std::move(x) returns an unnamed rvalue reference to x. rvalue references are really just like normal references. Their only purpose is simply to carry along the information about the "rvalue-ness" of the thing they refer to. When you then use the result of std::move() to initialize/assign to another object, overload resolution will pick a move constructor/move assignment operator if one exists. And that's it. That is literally all that std::move() does. However, the implementation of a move constructor/move assignment operator knows that the only way it could have been called is when the value passed to it is about to expire (otherwise, the copy constructor/copy assignment operator would have been called instead). It, thus, can safely "steal" the value rather than make a copy, whatever that may mean in the context of the particular type.
There is no general answer to the question what exactly it means to "steal" a value from an object. Whoever defines a type has to define whether it makes sense to move objects of this type and what exactly it means to do so (by declaring/defining the respective member functions). Built-in types don't have any special behavior defined for moving their values. So in the case of an int you just get what you get when you initialize an int with a reference to another int, which is a copy…
I made a two element Vector struct and I want to overload the + operator.
I made all my functions and methods take references, rather than values, and I want the + operator to work the same way.
impl Add for Vector {
fn add(&self, other: &Vector) -> Vector {
Vector {
x: self.x + other.x,
y: self.y + other.y,
}
}
}
Depending on which variation I try, I either get lifetime problems or type mismatches. Specifically, the &self argument seems to not get treated as the right type.
I have seen examples with template arguments on impl as well as Add, but they just result in different errors.
I found How can an operator be overloaded for different RHS types and return values? but the code in the answer doesn't work even if I put a use std::ops::Mul; at the top.
I am using rustc 1.0.0-nightly (ed530d7a3 2015-01-16 22:41:16 +0000)
I won't accept "you only have two fields, why use a reference" as an answer; what if I wanted a 100 element struct? I will accept an answer that demonstrates that even with a large struct I should be passing by value, if that is the case (I don't think it is, though.) I am interested in knowing a good rule of thumb for struct size and passing by value vs struct, but that is not the current question.
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 are types like any other1, so they can implement traits too. When a trait is implemented on a reference type, the type of self is a reference; the reference is passed by value. Normally, passing by value in Rust implies transferring ownership, but when references are passed by value, they're simply copied (or reborrowed/moved if it's a mutable reference), and that doesn't transfer ownership of the referent (because a reference doesn't own its referent in the first place). Considering all this, it makes sense for Add::add (and many other operators) to take self by value: if you need to take ownership of the operands, you can implement Add on structs/enums directly, and if you don't, you can implement Add on references.
Here, self is of type &'a Vector, because that's the type we're implementing Add on.
Note that I also specified the RHS type parameter with a different lifetime to emphasize the fact that the lifetimes of the two input parameters are unrelated.
1 Actually, reference types are special in that you can implement traits for references to types defined in your crate (i.e. if you're allowed to implement a trait for T, then you're also allowed to implement it for &T). &mut T and Box<T> have the same behavior, but that's not true in general for U<T> where U is not defined in the same crate.
If you want to support all scenarios, you must support all the combinations:
&T op U
T op &U
&T op &U
T op U
In rust proper, this was done through an internal macro.
Luckily, there is a rust crate, impl_ops, that also offers a macro to write that boilerplate for us: the crate offers the impl_op_ex! macro, which generates all the combinations.
Here is their sample:
#[macro_use] extern crate impl_ops;
use std::ops;
impl_op_ex!(+ |a: &DonkeyKong, b: &DonkeyKong| -> i32 { a.bananas + b.bananas });
fn main() {
let total_bananas = &DonkeyKong::new(2) + &DonkeyKong::new(4);
assert_eq!(6, total_bananas);
let total_bananas = &DonkeyKong::new(2) + DonkeyKong::new(4);
assert_eq!(6, total_bananas);
let total_bananas = DonkeyKong::new(2) + &DonkeyKong::new(4);
assert_eq!(6, total_bananas);
let total_bananas = DonkeyKong::new(2) + DonkeyKong::new(4);
assert_eq!(6, total_bananas);
}
Even better, they have a impl_op_ex_commutative! that'll also generate the operators with the parameters reversed if your operator happens to be commutative.
I was wondering if you know of any programming language in which we can pass parameters inside method name. I'm guessing this could improve the code readability. I.e.
Lets say I want to multiply to integers in a method. Normally my method declaration would be something like:
function multiply(int a, int b){
return a*b;
}
However, it may be nice to be able to define it this way also:
function multiply (int a) times (int b){
return a*b;
}
This way, we could make a more explicit call in the code by calling:
var c = multiply(4)times(2);
This could have a greater impact on more complicated code and programming syntax.
Do you know if something like this exists?
Of course, there is Smalltalk which is really expressive with its keyword messages...
n := collection size // 2.
head := collection copyFrom: 1 to: n.
Other than that, you will find that in ADA or Python, and probably some others you can prefix each argument with a key (which should match the function parameter names)...
I have a pointer to a parent class and I want to assign a new child object to that pointer conditionally. Right now, the syntax I have is rather lengthly:
std::unique_ptr<ParentClass> parentPtr;
if (...) {
parentPtr = std::unique_ptr<ParentClass>(new ChildClass1());
} else {
parentPtr = std::unique_ptr<ParentClass>(new ChildClass2());
}
Is there a good way of making this more readable / less lengthly?
Two possibilities would be:
std::unique_ptr<ParentClass> parentPtr(condition ?
(ParentClass*)new ChildClass1() :
(ParentClass*)new ChildClass2());
If condition is complicated, just assign a boolean to it and then write the construction. This solution only works for a binary condition though.
Another is to embrace C++14, and use
parentPtr = std::make_unique<ChildClass>();
First off, the "obvious" solution C ? new X : new Y does not work, since even if X and Y have a common base class A, the types X * and Y * have no common type. This is actually not so surprising after all if you consider that a class can have many bases (direct or indirect) and a given type may appear as a base multiple times.
You could make the conditional operator work by inserting a cast:
A * = C ? static_cast<A *>(new X) : static_cast<A *>(new Y);
But this would quickly get long and tedious to read when you try to apply this to your real situation.
However, as for std::unique_ptr, it offers the reset function which can be used to good effect here:
std::unique_ptr<A> p;
if (C)
{
p.reset(new X);
}
else
{
p.reset(new Y);
}
Now even if the actual new expressions are long, this is still nicely readable.
I'm trying to make a function that defines a vector that varies based on the function's input, and set! works great for this in Scheme. Is there a functional equivalent for this in OCaml?
I agree with sepp2k that you should expand your question, and give more detailed examples.
Maybe what you need are references.
As a rough approximation, you can see them as variables to which you can assign:
let a = ref 5;;
!a;; (* This evaluates to 5 *)
a := 42;;
!a;; (* This evaluates to 42 *)
Here is a more detailed explanation from http://caml.inria.fr/pub/docs/u3-ocaml/ocaml-core.html:
The language we have described so far is purely functional. That is, several evaluations of the same expression will always produce the same answer. This prevents, for instance, the implementation of a counter whose interface is a single function next : unit -> int that increments the counter and returns its new value. Repeated invocation of this function should return a sequence of consecutive integers — a different answer each time.
Indeed, the counter needs to memorize its state in some particular location, with read/write accesses, but before all, some information must be shared between two calls to next. The solution is to use mutable storage and interact with the store by so-called side effects.
In OCaml, the counter could be defined as follows:
let new_count =
let r = ref 0 in
let next () = r := !r+1; !r in
next;;
Another, maybe more concrete, example of mutable storage is a bank account. In OCaml, record fields can be declared mutable, so that new values can be assigned to them later. Hence, a bank account could be a two-field record, its number, and its balance, where the balance is mutable.
type account = { number : int; mutable balance : float }
let retrieve account requested =
let s = min account.balance requested in
account.balance <- account.balance -. s; s;;
In fact, in OCaml, references are not primitive: they are special cases of mutable records. For instance, one could define:
type 'a ref = { mutable content : 'a }
let ref x = { content = x }
let deref r = r.content
let assign r x = r.content <- x; x
set! in Scheme assigns to a variable. You cannot assign to a variable in OCaml, at all. (So "variables" are not really "variable".) So there is no equivalent.
But OCaml is not a pure functional language. It has mutable data structures. The following things can be assigned to:
Array elements
String elements
Mutable fields of records
Mutable fields of objects
In these situations, the <- syntax is used for assignment.
The ref type mentioned by #jrouquie is a simple, built-in mutable record type that acts as a mutable container of one thing. OCaml also provides ! and := operators for working with refs.