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+// Rust-101, Part 04: Ownership, Borrowing
+// =======================================
+
+use std::cmp;
+
+// Rust aims to be a "safe systems language". As a systems language, of course it
+// provides *references* (or *pointers*). But as a safe language, it has to
+// prevent bugs like this C++ snippet.
+/*
+ void foo(std::vector<int> v) {
+ int &first = v[0];
+ v.push_back(42);
+ first = 1337; // This is bad!
+ }
+*/
+// What's going wrong here? `first` is a reference into the vector `v`.
+// The operation `push_back` may re-allocate the storage for the vector,
+// in case the old buffer was full. If that happens, `first` is now
+// a dangling pointer, and accessing it can crash the program (or worse).
+//
+// It turns out that only the combination of two circumstances can lead to such a bug:
+// *aliasing* and *mutation*. In the code above, we have `first` and the buffer of `v`
+// being aliases, and when `push_back` is called, the latter is used to perform a mutation.
+// Therefore, the central principle of the Rust typesystem is to *rule out mutation in the presence
+// of aliasing*. The core tool to achieve that is the notion of *ownership*.
+
+// What does that mean in practice? Consider the following example.
+fn take(v: Vec<i32>) { /* do something */ }
+fn foo1() {
+ let v = vec![1,2,3,4];
+ take(v);
+ /* println!("The first element is: {}", v[0]); */
+}
+// Rust attaches additional meaning to the argument of `take`: The function can assume
+// that it entirely *owns* `v`, and hence can do anything with it. When `take` ends,
+// nobody needs `v` anymore, so it will be deleted (including its buffer on the heap).
+// Passing a `Vec<i32>` to `take` is considered *transfer of ownership*: Someone used
+// to own that vector, but now he gave it on to `take` and has no business with it anymore.
+//
+// If you give a book to your friend, you cannot some to his place next day and get the book!
+// It's no longer yours. Rust makes sure you don't break this rule. Try enabling the commented
+// line in `foo1`. Rust will tell you that `v` has been *moved*, which is to say that ownership
+// has been transferred somewhere else. In this particular case, the buffer storing the data
+// does not even exist anymore, so we are lucky that Rust caught this problem!
+// Essentially, ownership rules out aliasing, hence making the kind of problem discussed above
+// impossible.
+
+// If you go back to our example with `vec_min`, and try to call that function twice, you will
+// get the same error. That's because `vec_min` demands that the caller transfers ownership of the
+// vector. Hence, when `vec_min` finishes, the entire vector is deleted. That's of course not what
+// we wanted! Can't we somehow give `vec_min` access to the vector, while retaining ownership of it?
+//
+// Rust calls this *borrowing* the vector, and it works a bit like borrowing does in the real world:
+// If you borrow a book to your friend, your friend can have it and work on it (and you can't!)
+// as long as the book is still borrowed. Your friend could even borrow the book to someone else.
+// Eventually however, your friend has to give the book back to you, at which point you again
+// have full control.
+//
+// Rust distinguishes between two kinds of borrows. First of all, there's the *shared* borrow.
+// This is where the book metaphor kind of breaks down... you can give a shared borrow of
+// *the same data* to lots of different people, who can all access the data. This of course
+// introduces aliasing, so in order to live up to its promise of safety, Rust does not allow
+// mutation through a shared borrow.
+
+// So, let's re-write `vec_min` to work on a shared borrow of a vector. In fact, the only
+// thing we have to change is the type of the function. The `e` in the loop now gets type
+// `&i32`, hence we have to deference it.
+fn vec_min(v: &Vec<i32>) -> Option<i32> {
+ let mut min = None;
+ for e in v {
+ min = Some(match min {
+ None => *e,
+ Some(n) => cmp::min(n, *e)
+ });
+ }
+ min
+}
+
+// Now that `vec_min` does not acquire ownership of the vector anymore, we can call it multiple times on the same vector and also do things like
+fn foo2() {
+ let v = vec![5,4,3,2,1];
+ let first = &v[0];
+ vec_min(&v);
+ vec_min(&v);
+ println!("The first element is: {}", *first);
+}
+// What's going on here? First, `&` is how you create a shared borrow to something. This code creates three
+// shared borrows to `v`: The borrow for `first` begins in the 2nd line of the function and lasts all the way to
+// the end. The other two borrows, created for calling `vec_min`, only last for the duration of that
+// call.
+//
+// Technically, of course, borrows are pointers. Notice that since `vec_min` only gets a shared
+// borrow, Rust knows that it cannot mutate `v` in any way. Hence the pointer created before calling
+// `vec_min` remains valid.
+
+// There is a second kind of borrow, a *mutable borrow*. As the name suggests, such a borrow permits
+// mutation, and hence has to prevent aliasing. There can only ever be one mutable borrow to a
+// particular piece of data.
+
+// As an example, consider a function which increments every element of a vector by 1.
+fn vec_inc(v: &mut Vec<i32>) {
+ for e in v {
+ *e += 1;
+ }
+}
+// The type `&mut Vec<i32>` is the type of mutable borrows of `vec<i32>`. Because the borrow is
+// mutable, we can change `e` in the loop. How can we call this function?
+fn foo3() {
+ let mut v = vec![5,4,3,2,1];
+ /* let first = &v[0]; */
+ vec_inc(&mut v);
+ vec_inc(&mut v);
+ /* println!("The first element is: {}", *first); */
+}
+// `&mut` is the operator to create a mutable borrow. We have to mark `v` as mutable in order
+// to create such a borrow. Because the borrow passed to `vec_inc` only lasts as
+// long as the function call, we can still call `vec_inc` on the same vector twice:
+// The durations of the two borrows do not overlap, so we never have more than one mutable borrow.
+// However, we can *not* create a shared borrow that spans a call to `vec_inc`. Just try
+// enabling the commented-out lines. This is because `vec_inc` could mutate the vector structurally
+// (i.e., it could add or remove elements), and hence the pointer `first` could become invalid.
+//
+// Above, I said that having a mutable borrow excludes aliasing. But if you look at the code above carefully,
+// you may say: "Wait! Don't the `v` in `foo3` and the `v` in `vec_inc` alias?" And you are right,
+// they do. However, the `v` in `foo3` is not actually usable, it is not *active*: As long as there is an
+// outstanding borrow, Rust will not allow you to do anything with `v`. This is, in fact, what
+// prevents the creation of a mutable borrow when there already is a shared one.
+
+// This also works the other way around: In `foo4`, there is already a mutable borrow active in the `vec_min`
+// line, so the attempt to create another shared borrow is rejected by the compiler.
+fn foo4() {
+ let mut v = vec![5,4,3,2,1];
+ let first = &mut v[0];
+ /* vec_min(&v); */
+ println!("The first element is: {}", *first);
+}
+
+// So, to summarize: The ownership and borrowing system of Rust enforces the following three rules:
+//
+// * There is always exactly one owner of a piece of data
+// * If there is an active mutable borrow, then nobody else can have active access to the data
+// * If there is an active shared borrow, then every other active access to the data is also a shared borrow
+//
+// As it turns out, combined with the abstraction facilities of Rust, this is a very powerful mechanism
+// to tackle many problems beyond basic memory safety.
+
+// [index](main.html) | [previous](part03.html) | [next](main.html)
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