I am new to rust. My simple code
struct Foo<'a>{
data : & 'a String,
}
fn test_foo(){
let s1:String = String::from("A");
let mut foo:Foo = Foo { data: & s1 };
let s2:String = String::from("B");
foo.data = & s2;
}
My query : If I implement Drop trait for Foo struct
foo.data = & s2;
starts giving compilation error : s2 does not live long enough. The error makes sense to me. But, why am I not getting the same error if I don't implement Drop trait for Foo
In Rust, local variables (from the same scope) are dropped in reverse order of the order they're defined.
fn test_foo(){
let s1:String = String::from("A");
let mut foo:Foo = Foo { data: & s1 };
let s2:String = String::from("B");
foo.data = & s2;
}
We declare three local variables here: s1, foo, and s2, in that order. Rust would like to drop them in reverse order: s2 first, then foo, then s1. However, there's a problem with the lifetimes. Once we drop s2, then foo.data is uninitialized, i.e. it points to garbage memory.
Now, why does this work without Drop? Rust has a concept called partial moves. If you have a structure that has multiple fields, Rust will allow you to move out of some fields without invalidating the whole struct. In principle, if I have a
struct Person {
name: String,
age: i32,
occupation: String,
}
and I do let name = my_person.name; (where my_person: Person), then I've moved a value out of a Person. Thus, my_person.name is invalid, and in the absence of partial moves, Rust should consider my_person completely invalid. However, we know that my_person.age and my_person.occupation are still valid, so Rust will let age and occupation remain in place. It remembers that name is moved (and hence garbage that should not be dropped), while age and occupation are still valid.
In your example, the same thing is happening. Rust wants to drop s2, but Foo still holds a reference to it. Rust considers that fine: We'll drop s2 and simply say that Foo has been partially moved: Its data field is no longer valid. Then when we go to drop foo next, we don't need to drop the reference, merely the outermost Foo layer itself.
Without a Drop instance, this is fine, and Rust will allow it. However, if impl<'a> Drop for Foo<'a> is in scope, then partial moves are completely disabled for Foo. Rust sees that you're implementing some custom Drop behavior, and now it won't allow a partially-initialized object to exist, since we would have to drop a partially-initialized object, and Rust can't predict what your custom Drop code is going to do or what assumptions it's going to make.
So with a Drop implementation, Rust still wants to drop s2 first, but it can't partially move the reference out of foo, since that would leave foo in a partially-initialized state, which isn't allowed.
Related
I'm writing my first rust program and as expected I'm having problems making the borrow checker happy. Here is what I'm trying to do:
I would like to have a function that allocates some array, stores the array in some global data structure, and returns a reference to it. Example:
static mut global_data = ...
fn f() -> &str {
let s = String::new();
global.my_string = s;
return &s;
};
Is there any way to make something like this work? If not, what is "the rust way"(tm) to get an array and a pointer into it?
Alternatively, is there any documentation I could read? The rust book is unfortunately very superficial on most topics.
There are a couple things wrong with your code:
Using global state is very unidiomatic in rust. It can be done in some specific scenarios, but it should never be a go to method. You cold try wrapping your state in Rc or Arc and share it this way in your program. If you also want to mutate this state (as you show in your example) you must to wrap it also in some kind of interior mutability type. So try Rc<RefCell<State>> if you want to use state in only one thread or Arc<Mutex<State>> if you want to use it from multiple different threads.
Accessing mutable static memory is unsafe. So even the following code won't compile:
static mut x: i32 = 0;
// neither of this lines work!
println!("{}", x);
x = 42;
You must use unsafe to access or modify any static mutable variables, because you must de facto prove to the compiler that you assure it that no data races (from accessing this data from different threads) will occur.
I can't be sure, since you didn't show what type is global_data, but I assume, that my_string is a field of type String. When you write
let s = String::new();
global.my_string = s;
You move ownership of that string to the global. You therefore cannot return (or even create) reference to it. You must do this though it's new owner. &global.my_string could work, but not if you do what I written in 1. You could try to return RefMut of MutexGuard, but that is probably not what you want.
Okay, just in case someone else is having the same question, the following code seems to work:
struct foo {
b : Option<Box<u32>>,
}
static mut global : foo = foo { b : None };
fn f<'a>() -> &'a u32 {
let b : Box<u32> = Box::new(5);
unsafe {
global.b = Some(b);
match &global.b {
None => panic!(""),
Some(a) => return &a,
}
}
}
At least it compiles. Hopefully it will also do the right thing when run.
I'm aware that this is not how you are supposed to do things in rust. But I'm currently trying to figure out how to implement various data structures from scratch, and the above is just a reduced example of one of the problems I encountered.
The Rust language website claims move semantics as one of the features of the language. But I can't see how move semantics is implemented in Rust.
Rust boxes are the only place where move semantics are used.
let x = Box::new(5);
let y: Box<i32> = x; // x is 'moved'
The above Rust code can be written in C++ as
auto x = std::make_unique<int>(5);
auto y = std::move(x); // Note the explicit move
As far as I know (correct me if I'm wrong),
Rust doesn't have constructors at all, let alone move constructors.
No support for rvalue references.
No way to create functions overloads with rvalue parameters.
How does Rust provide move semantics?
I think it's a very common issue when coming from C++. In C++ you are doing everything explicitly when it comes to copying and moving. The language was designed around copying and references. With C++11 the ability to "move" stuff was glued onto that system. Rust on the other hand took a fresh start.
Rust doesn't have constructors at all, let alone move constructors.
You do not need move constructors. Rust moves everything that "does not have a copy constructor", a.k.a. "does not implement the Copy trait".
struct A;
fn test() {
let a = A;
let b = a;
let c = a; // error, a is moved
}
Rust's default constructor is (by convention) simply an associated function called new:
struct A(i32);
impl A {
fn new() -> A {
A(5)
}
}
More complex constructors should have more expressive names. This is the named constructor idiom in C++
No support for rvalue references.
It has always been a requested feature, see RFC issue 998, but most likely you are asking for a different feature: moving stuff to functions:
struct A;
fn move_to(a: A) {
// a is moved into here, you own it now.
}
fn test() {
let a = A;
move_to(a);
let c = a; // error, a is moved
}
No way to create functions overloads with rvalue parameters.
You can do that with traits.
trait Ref {
fn test(&self);
}
trait Move {
fn test(self);
}
struct A;
impl Ref for A {
fn test(&self) {
println!("by ref");
}
}
impl Move for A {
fn test(self) {
println!("by value");
}
}
fn main() {
let a = A;
(&a).test(); // prints "by ref"
a.test(); // prints "by value"
}
Rust's moving and copying semantics are very different from C++. I'm going to take a different approach to explain them than the existing answer.
In C++, copying is an operation that can be arbitrarily complex, due to custom copy constructors. Rust doesn't want custom semantics of simple assignment or argument passing, and so takes a different approach.
First, an assignment or argument passing in Rust is always just a simple memory copy.
let foo = bar; // copies the bytes of bar to the location of foo (might be elided)
function(foo); // copies the bytes of foo to the parameter location (might be elided)
But what if the object controls some resources? Let's say we are dealing with a simple smart pointer, Box.
let b1 = Box::new(42);
let b2 = b1;
At this point, if just the bytes are copied over, wouldn't the destructor (drop in Rust) be called for each object, thus freeing the same pointer twice and causing undefined behavior?
The answer is that Rust moves by default. This means that it copies the bytes to the new location, and the old object is then gone. It is a compile error to access b1 after the second line above. And the destructor is not called for it. The value was moved to b2, and b1 might as well not exist anymore.
This is how move semantics work in Rust. The bytes are copied over, and the old object is gone.
In some discussions about C++'s move semantics, Rust's way was called "destructive move". There have been proposals to add the "move destructor" or something similar to C++ so that it can have the same semantics. But move semantics as they are implemented in C++ don't do this. The old object is left behind, and its destructor is still called. Therefore, you need a move constructor to deal with the custom logic required by the move operation. Moving is just a specialized constructor/assignment operator that is expected to behave in a certain way.
So by default, Rust's assignment moves the object, making the old location invalid. But many types (integers, floating points, shared references) have semantics where copying the bytes is a perfectly valid way of creating a real copy, with no need to ignore the old object. Such types should implement the Copy trait, which can be derived by the compiler automatically.
#[derive(Copy)]
struct JustTwoInts {
one: i32,
two: i32,
}
This signals the compiler that assignment and argument passing do not invalidate the old object:
let j1 = JustTwoInts { one: 1, two: 2 };
let j2 = j1;
println!("Still allowed: {}", j1.one);
Note that trivial copying and the need for destruction are mutually exclusive; a type that is Copy cannot also be Drop.
Now what about when you want to make a copy of something where just copying the bytes isn't enough, e.g. a vector? There is no language feature for this; technically, the type just needs a function that returns a new object that was created the right way. But by convention this is achieved by implementing the Clone trait and its clone function. In fact, the compiler supports automatic derivation of Clone too, where it simply clones every field.
#[Derive(Clone)]
struct JustTwoVecs {
one: Vec<i32>,
two: Vec<i32>,
}
let j1 = JustTwoVecs { one: vec![1], two: vec![2, 2] };
let j2 = j1.clone();
And whenever you derive Copy, you should also derive Clone, because containers like Vec use it internally when they are cloned themselves.
#[derive(Copy, Clone)]
struct JustTwoInts { /* as before */ }
Now, are there any downsides to this? Yes, in fact there is one rather big downside: because moving an object to another memory location is just done by copying bytes, and no custom logic, a type cannot have references into itself. In fact, Rust's lifetime system makes it impossible to construct such types safely.
But in my opinion, the trade-off is worth it.
Rust supports move semantics with features like these:
All types are moveable.
Sending a value somewhere is a move, by default, throughout the language. For non-Copy types, like Vec, the following are all moves in Rust: passing an argument by value, returning a value, assignment, pattern-matching by value.
You don't have std::move in Rust because it's the default. You're really using moves all the time.
Rust knows that moved values must not be used. If you have a value x: String and do channel.send(x), sending the value to another thread, the compiler knows that x has been moved. Trying to use it after the move is a compile-time error, "use of moved value". And you can't move a value if anyone has a reference to it (a dangling pointer).
Rust knows not to call destructors on moved values. Moving a value transfers ownership, including responsibility for cleanup. Types don't have to be able to represent a special "value was moved" state.
Moves are cheap and the performance is predictable. It's basically memcpy. Returning a huge Vec is always fast—you're just copying three words.
The Rust standard library uses and supports moves everywhere. I already mentioned channels, which use move semantics to safely transfer ownership of values across threads. Other nice touches: all types support copy-free std::mem::swap() in Rust; the Into and From standard conversion traits are by-value; Vec and other collections have .drain() and .into_iter() methods so you can smash one data structure, move all the values out of it, and use those values to build a new one.
Rust doesn't have move references, but moves are a powerful and central concept in Rust, providing a lot of the same performance benefits as in C++, and some other benefits as well.
let s = vec!["udon".to_string(), "ramen".to_string(), "soba".to_string()];
this is how it is represented in memory
Then let's assign s to t
let t = s;
this is what happens:
let t = s MOVED the vector’s three header fields from s to t; now t is the owner of the vector. The vector’s elements stayed just
where they were, and nothing happened to the strings either. Every value still has a single owner.
Now s is freed, if I write this
let u = s
I get error: "use of moved value: s"
Rust applies move semantics to almost any use of a value (Except Copy types). Passing
arguments to functions moves ownership to the function’s parameters;
returning a value from a function moves ownership to the caller.
Building a tuple moves the values into the tuple. And so on.
Ref for example:Programming Rust by Jim Blandy, Jason Orendorff, Leonora F. S. Tindall
Primitive types cannot be empty and are fixed size while non primitives can grow and can be empty. since primitive types cannot be empty and are fixed size, therefore assigning memory to store them and handling them are relatively easy. however the handling of non primitives involves the computation of how much memory they will take as they grow and other costly operations.Wwith primitives rust will make a copy, with non primitive rust does a move
fn main(){
// this variable is stored in stack. primitive types are fixed size, we can store them on stack
let x:i32=10;
// s1 is stored in heap. os will assign memory for this. pointer of this memory will be stored inside stack.
// s1 is the owner of memory space in heap which stores "my name"
// if we dont clear this memory, os will have no access to this memory. rust uses ownership to free the memory
let s1=String::from("my name");
// s1 will be cleared from the stack, s2 will be added to the stack poniting the same heap memory location
// making new copy of this string will create extra overhead, so we MOVED the ownership of s1 into s2
let s2=s1;
// s3 is the pointer to s2 which points to heap memory. we Borrowed the ownership
// Borrowing is similar borrowing in real life, you borrow a car from your friend, but its ownership does not change
let s3=&s2;
// this is creating new "my name" in heap and s4 stored as the pointer of this memory location on the heap
let s4=s2.clone()
}
Same principle applies when we pass primitive or non-primitive type arguments to a function:
fn main(){
// since this is primitive stack_function will make copy of it so this will remain unchanged
let stack_num=50;
let mut heap_vec=vec![2,3,4];
// when we pass a stack variable to a function, function will make a copy of that and will use the copy. "move" does not occur here
stack_var_fn(stack_num);
println!("The stack_num inside the main fn did not change:{}",stack_num);
// the owner of heap_vec moved here and when function gets executed, it goes out of scope so the variable will be dropped
// we can pass a reference to reach the value in heap. so we use the pointer of heap_vec
// we use "&"" operator to indicate that we are passing a reference
heap_var_fn(&heap_vec);
println!("the heap_vec inside main is:{:?}",heap_vec);
}
// this fn that we pass an argument stored in stack
fn stack_var_fn(mut var:i32){
// we are changing the arguments value
var=56;
println!("Var inside stack_var_fn is :{}",var);
}
// this fn that we pass an arg that stored in heap
fn heap_var_fn(var:&Vec<i32>){
println!("Var:{:?}",var);
}
I would like to add that it is not necessary for move to memcpy. If the object on the stack is large enough, Rust's compiler may choose to pass the object's pointer instead.
In C++ the default assignment of classes and structs is shallow copy. The values are copied, but not the data referenced by pointers. So modifying one instance changes the referenced data of all copies. The values (f.e. used for administration) remain unchanged in the other instance, likely rendering an inconsistent state. A move semantic avoids this situation. Example for a C++ implementation of a memory managed container with move semantic:
template <typename T>
class object
{
T *p;
public:
object()
{
p=new T;
}
~object()
{
if (p != (T *)0) delete p;
}
template <typename V> //type V is used to allow for conversions between reference and value
object(object<V> &v) //copy constructor with move semantic
{
p = v.p; //move ownership
v.p = (T *)0; //make sure it does not get deleted
}
object &operator=(object<T> &v) //move assignment
{
delete p;
p = v.p;
v.p = (T *)0;
return *this;
}
T &operator*() { return *p; } //reference to object *d
T *operator->() { return p; } //pointer to object data d->
};
Such an object is automatically garbage collected and can be returned from functions to the calling program. It is extremely efficient and does the same as Rust does:
object<somestruct> somefn() //function returning an object
{
object<somestruct> a;
auto b=a; //move semantic; b becomes invalid
return b; //this moves the object to the caller
}
auto c=somefn();
//now c owns the data; memory is freed after leaving the scope
To illustrate the necessity of Rc<T>, the Book presents the following snippet (spoiler: it won't compile) to show that we cannot enable multiple ownership without Rc<T>.
enum List {
Cons(i32, Box<List>),
Nil,
}
use crate::List::{Cons, Nil};
fn main() {
let a = Cons(5, Box::new(Cons(10, Box::new(Nil))));
let b = Cons(3, Box::new(a));
let c = Cons(4, Box::new(a));
}
It then claims (emphasis mine)
We could change the definition of Cons to hold references instead, but then we would have to specify lifetime parameters. By specifying lifetime parameters, we would be specifying that every element in the list will live at least as long as the entire list. The borrow checker wouldn’t let us compile let a = Cons(10, &Nil); for example, because the temporary Nil value would be dropped before a could take a reference to it.
Well, not quite. The following snippet compiles under rustc 1.52.1
enum List<'a> {
Cons(i32, &'a List<'a>),
Nil,
}
use crate::List::{Cons, Nil};
fn main() {
let a = Cons(5, &Cons(10, &Nil));
let b = Cons(3, &a);
let c = Cons(4, &a);
}
Note that by taking a reference, we no longer need a Box<T> indirection to hold the nested List. Furthermore, I can point both b and c to a, which gives a multiple conceptual owners (which are actually borrowers).
Question: why do we need Rc<T> when immutable references can do the job?
With "ordinary" borrows you can very roughly think of a statically proven order-by-relationship, where the compiler needs to prove that the owner of something always comes to life before any borrows and always dies after all borrows died (a owns String, it comes to life before b which borrows a, then b dies, then a dies; valid). For a lot of use-cases, this can be done, which is Rust's insight to make the borrow-system practical.
There are cases where this can't be done statically. In the example you've given, you're sort of cheating, because all borrows have a 'static-lifetime; and 'static items can be "ordered" before or after anything out to infinity because of that - so there actually is no constraint in the first place. The example becomes much more complex when you take different lifetimes (many List<'a>, List<'b>, etc.) into account. This issue will become apparent when you try to pass values into functions and those functions try to add items. This is because values created inside functions will die after leaving their scope (i.e. when the enclosing function returns), so we cannot keep a reference to them afterwards, or there will be dangling references.
Rc comes in when one can't prove statically who is the original owner, whose lifetime starts before any other and ends after any other(!). A classic example is a graph structure derived from user input, where multiple nodes can refer to one other node. They need to form a "born after, dies before" relationship with the node they are referencing at runtime, to guarantee that they never reference invalid data. The Rc is a very simple solution to that because a simple counter can represent these relationships. As long as the counter is not zero, some "born after, dies before" relationship is still active. The key insight here is that it does not matter in which order the nodes are created and die because any order is valid. Only the points on either end - where the counter gets to 0 - are actually important, any increase or decrease in between is the same (0=+1+1+1-1-1-1=0 is the same as 0=+1+1-1+1-1-1=0) The Rc is destroyed when the counter reaches zero. In the graph example this is when a node is not being referred to any longer. This tells the owner of that Rc (the last node referring) "Oh, it turns out I am the owner of the underlying node - nobody knew! - and I get to destroy it".
Even single-threaded, there are still times the destruction order is determined dynamically, whereas for the borrow checker to work, there must be a determined lifetime tree (stack).
fn run() {
let writer = Rc::new(std::io::sink());
let mut counters = vec![
(7, Rc::clone(&writer)),
(7, writer),
];
while !counters.is_empty() {
let idx = read_counter_index();
counters[idx].0 -= 1;
if counters[idx].0 == 0 {
counters.remove(idx);
}
}
}
fn read_counter_index() -> usize {
unimplemented!()
}
As you can see in this example, the order of destruction is determined by user input.
Another reason to use smart pointers is simplicity. The borrow checker does incur some code complexity. For example, using smart pointer, you are able to maneuver around the self-referential struct problem with a tiny overhead.
struct SelfRefButDynamic {
a: Rc<u32>,
b: Rc<u32>,
}
impl SelfRefButDynamic {
pub fn new() -> Self {
let a = Rc::new(0);
let b = Rc::clone(&a);
Self { a, b }
}
}
This is not possible with static (compile-time) references:
struct WontDo {
a: u32,
b: &u32,
}
In Rust, there are two possibilities to take a reference
Borrow, i.e., take a reference but don't allow mutating the reference destination. The & operator borrows ownership from a value.
Borrow mutably, i.e., take a reference to mutate the destination. The &mut operator mutably borrows ownership from a value.
The Rust documentation about borrowing rules says:
First, any borrow must last for a scope no greater than that of the
owner. Second, you may have one or the other of these two kinds of
borrows, but not both at the same time:
one or more references (&T) to a resource,
exactly one mutable reference (&mut T).
I believe that taking a reference is creating a pointer to the value and accessing the value by the pointer. This could be optimized away by the compiler if there is a simpler equivalent implementation.
However, I don't understand what move means and how it is implemented.
For types implementing the Copy trait it means copying e.g. by assigning the struct member-wise from the source, or a memcpy(). For small structs or for primitives this copy is efficient.
And for move?
This question is not a duplicate of What are move semantics? because Rust and C++ are different languages and move semantics are different between the two.
Semantics
Rust implements what is known as an Affine Type System:
Affine types are a version of linear types imposing weaker constraints, corresponding to affine logic. An affine resource can only be used once, while a linear one must be used once.
Types that are not Copy, and are thus moved, are Affine Types: you may use them either once or never, nothing else.
Rust qualifies this as a transfer of ownership in its Ownership-centric view of the world (*).
(*) Some of the people working on Rust are much more qualified than I am in CS, and they knowingly implemented an Affine Type System; however contrary to Haskell which exposes the math-y/cs-y concepts, Rust tends to expose more pragmatic concepts.
Note: it could be argued that Affine Types returned from a function tagged with #[must_use] are actually Linear Types from my reading.
Implementation
It depends. Please keep in mind than Rust is a language built for speed, and there are numerous optimizations passes at play here which will depend on the compiler used (rustc + LLVM, in our case).
Within a function body (playground):
fn main() {
let s = "Hello, World!".to_string();
let t = s;
println!("{}", t);
}
If you check the LLVM IR (in Debug), you'll see:
%_5 = alloca %"alloc::string::String", align 8
%t = alloca %"alloc::string::String", align 8
%s = alloca %"alloc::string::String", align 8
%0 = bitcast %"alloc::string::String"* %s to i8*
%1 = bitcast %"alloc::string::String"* %_5 to i8*
call void #llvm.memcpy.p0i8.p0i8.i64(i8* %1, i8* %0, i64 24, i32 8, i1 false)
%2 = bitcast %"alloc::string::String"* %_5 to i8*
%3 = bitcast %"alloc::string::String"* %t to i8*
call void #llvm.memcpy.p0i8.p0i8.i64(i8* %3, i8* %2, i64 24, i32 8, i1 false)
Underneath the covers, rustc invokes a memcpy from the result of "Hello, World!".to_string() to s and then to t. While it might seem inefficient, checking the same IR in Release mode you will realize that LLVM has completely elided the copies (realizing that s was unused).
The same situation occurs when calling a function: in theory you "move" the object into the function stack frame, however in practice if the object is large the rustc compiler might switch to passing a pointer instead.
Another situation is returning from a function, but even then the compiler might apply "return value optimization" and build directly in the caller's stack frame -- that is, the caller passes a pointer into which to write the return value, which is used without intermediary storage.
The ownership/borrowing constraints of Rust enable optimizations that are difficult to reach in C++ (which also has RVO but cannot apply it in as many cases).
So, the digest version:
moving large objects is inefficient, but there are a number of optimizations at play that might elide the move altogether
moving involves a memcpy of std::mem::size_of::<T>() bytes, so moving a large String is efficient because it only copies a couple bytes whatever the size of the allocated buffer they hold onto
When you move an item, you are transferring ownership of that item. That's a key component of Rust.
Let's say I had a struct, and then I assign the struct from one variable to another. By default, this will be a move, and I've transferred ownership. The compiler will track this change of ownership and prevent me from using the old variable any more:
pub struct Foo {
value: u8,
}
fn main() {
let foo = Foo { value: 42 };
let bar = foo;
println!("{}", foo.value); // error: use of moved value: `foo.value`
println!("{}", bar.value);
}
how it is implemented.
Conceptually, moving something doesn't need to do anything. In the example above, there wouldn't be a reason to actually allocate space somewhere and then move the allocated data when I assign to a different variable. I don't actually know what the compiler does, and it probably changes based on the level of optimization.
For practical purposes though, you can think that when you move something, the bits representing that item are duplicated as if via memcpy. This helps explain what happens when you pass a variable to a function that consumes it, or when you return a value from a function (again, the optimizer can do other things to make it efficient, this is just conceptually):
// Ownership is transferred from the caller to the callee
fn do_something_with_foo(foo: Foo) {}
// Ownership is transferred from the callee to the caller
fn make_a_foo() -> Foo { Foo { value: 42 } }
"But wait!", you say, "memcpy only comes into play with types implementing Copy!". This is mostly true, but the big difference is that when a type implements Copy, both the source and the destination are valid to use after the copy!
One way of thinking of move semantics is the same as copy semantics, but with the added restriction that the thing being moved from is no longer a valid item to use.
However, it's often easier to think of it the other way: The most basic thing that you can do is to move / give ownership away, and the ability to copy something is an additional privilege. That's the way that Rust models it.
This is a tough question for me! After using Rust for a while the move semantics are natural. Let me know what parts I've left out or explained poorly.
Rust's move keyword always bothers me so, I decided to write my understanding which I obtained after discussion with my colleagues.
I hope this might help someone.
let x = 1;
In the above statement, x is a variable whose value is 1. Now,
let y = || println!("y is a variable whose value is a closure");
So, move keyword is used to transfer the ownership of a variable to the closure.
In the below example, without move, x is not owned by the closure. Hence x is not owned by y and available for further use.
let x = 1;
let y = || println!("this is a closure that prints x = {}". x);
On the other hand, in this next below case, the x is owned by the closure. x is owned by y and not available for further use.
let x = 1;
let y = move || println!("this is a closure that prints x = {}". x);
By owning I mean containing as a member variable. The example cases above are in the same situation as the following two cases. We can also assume the below explanation as to how the Rust compiler expands the above cases.
The formar (without move; i.e. no transfer of ownership),
struct ClosureObject {
x: &u32
}
let x = 1;
let y = ClosureObject {
x: &x
};
The later (with move; i.e. transfer of ownership),
struct ClosureObject {
x: u32
}
let x = 1;
let y = ClosureObject {
x: x
};
Please let me answer my own question. I had trouble, but by asking a question here I did Rubber Duck Problem Solving. Now I understand:
A move is a transfer of ownership of the value.
For example the assignment let x = a; transfers ownership: At first a owned the value. After the let it's x who owns the value. Rust forbids to use a thereafter.
In fact, if you do println!("a: {:?}", a); after the letthe Rust compiler says:
error: use of moved value: `a`
println!("a: {:?}", a);
^
Complete example:
#[derive(Debug)]
struct Example { member: i32 }
fn main() {
let a = Example { member: 42 }; // A struct is moved
let x = a;
println!("a: {:?}", a);
println!("x: {:?}", x);
}
And what does this move mean?
It seems that the concept comes from C++11. A document about C++ move semantics says:
From a client code point of view, choosing move instead of copy means that you don't care what happens to the state of the source.
Aha. C++11 does not care what happens with source. So in this vein, Rust is free to decide to forbid to use the source after a move.
And how it is implemented?
I don't know. But I can imagine that Rust does literally nothing. x is just a different name for the same value. Names usually are compiled away (except of course debugging symbols). So it's the same machine code whether the binding has the name a or x.
It seems C++ does the same in copy constructor elision.
Doing nothing is the most efficient possible.
Passing a value to function, also results in transfer of ownership; it is very similar to other examples:
struct Example { member: i32 }
fn take(ex: Example) {
// 2) Now ex is pointing to the data a was pointing to in main
println!("a.member: {}", ex.member)
// 3) When ex goes of of scope so as the access to the data it
// was pointing to. So Rust frees that memory.
}
fn main() {
let a = Example { member: 42 };
take(a); // 1) The ownership is transfered to the function take
// 4) We can no longer use a to access the data it pointed to
println!("a.member: {}", a.member);
}
Hence the expected error:
post_test_7.rs:12:30: 12:38 error: use of moved value: `a.member`
let s1:String= String::from("hello");
let s2:String= s1;
To ensure memory safety, rust invalidates s1, so instead of being shallow copy, this called a Move
fn main() {
// Each value in rust has a variable that is called its owner
// There can only be one owner at a time.
let s=String::from('hello')
take_ownership(s)
println!("{}",s)
// Error: borrow of moved value "s". value borrowed here after move. so s cannot be borrowed after a move
// when we pass a parameter into a function it is the same as if we were to assign s to another variable. Passing 's' moves s into the 'my_string' variable then `println!("{}",my_string)` executed, "my_string" printed out. After this scope is done, some_string gets dropped.
let x:i32 = 2;
makes_copy(x)
// instead of being moved, integers are copied. we can still use "x" after the function
//Primitives types are Copy and they are stored in stack because there size is known at compile time.
println("{}",x)
}
fn take_ownership(my_string:String){
println!('{}',my_string);
}
fn makes_copy(some_integer:i32){
println!("{}", some_integer)
}
I have the following simple program
fn main() {
let a = 10;
let b: i32;
let r: &i32;
b = a; // move?
r = &a; // borrow?
println!("{}", a);
println!("{}", b);
println!("{}", r);
println!("{}", &r);
println!("{}", *r);
}
The output is
10
10
10
10
10
The first print does not fail even when the value is moved. Is this because of primitive type or am I missing something?
The second print seems ok.
The third one prints a reference directly - shouldn't we get the memory address as this is a reference?
The fourth print is a reference to a reference, which should print a memory address, I think?
The fifth print seems reasonable as (I think) * is the value at operator that de-references the reference.
It seems I am not quite getting the whole thing.
Please explain in detail what's going on.
Related:
Move vs Copy in Rust
1, 2 => You are working with i32, which is Copy, so in practice b = a.clone()
3, 4, 5 => You're confused with the Deref trait. I find it easier to reason about ownership/borrowing than references in rust. r = &a means r borrows a so I can access its value later on, someone else will own it and take care of dropping it
Regarding 1: Yes, because it's a primitive variable, more specifically a type that implements the Copy trait. All those Copy-types work with copy semantics instead of move semantics.
Regarding 3: println! automatically dereferences it's arguments -- this is what the user wants in 99% of all cases.
Regarding 4: Again, automatically dereferences arguments... until it's a non-reference type.
The other answers are mostly right, but have some small errors.
1. i32 implements Copy, so when you assign it to a second variable binding, the first binding does not need to be invalidated. Any type that implements Copy will have this property.
3. You have asked to format the value with {} which corresponds to the Display trait. There is an implementation of this trait for references to types that implement Display:
impl<'a, T> Display for &'a T where T: Display + ?Sized {
fn fmt(&self, f: &mut Formatter) -> Result { Display::fmt(&**self, f) }
}
This simply delegates to the implementation of the referred-to type.
4. The same as #3 - a reference to a reference to a type that implements Display will just delegate twice. Deref does not come into play.
Here's the sneaky thing that no one else has mentioned. println! is a macro, which means it has more power than a regular function call. One of the things that it does is automatically take a reference to any arguments. That's what allows you to print out a value that doesn't implement Copy without losing ownership.
With this code:
let a = 10;
println!("{}", a);
The expanded version is actually something like this (slightly cleaned up):
let a = 10;
static __STATIC_FMTSTR: &'static [&'static str] = &["", "\n"];
::std::io::_print(::std::fmt::Arguments::new_v1(__STATIC_FMTSTR, &match (&a,) {
(__arg0,) => [::std::fmt::ArgumentV1::new(__arg0, ::std::fmt::Display::fmt)],
}));
Therefore, everything passed to println! is a reference. It wouldn't be very useful if references printed out memory addresses.
Besides the usefulness, Rust focuses more on value semantics as opposed to reference semantics. When you have values moving and changing addresses frequently, the location of the value isn't very consistent or useful.
See also
Auto-dereference when printing a pointer, or did I miss something?
Reference to a vector still prints as a vector?