// prevent bugs like this C++ snippet.
/*
void foo(std::vector<int> v) {
- int &first = v[0];
+ int *first = &v[0];
v.push_back(42);
- first = 1337; // This is bad!
+ *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).
+// What's going wrong here? `first` is a pointer 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`
// 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*.
+// ## Ownership
// What does that mean in practice? Consider the following example.
-fn take(v: Vec<i32>) { /* do something */ }
-fn foo1() {
+fn work_on_vector(v: Vec<i32>) { /* do something */ }
+fn ownership_demo() {
let v = vec![1,2,3,4];
- take(v);
+ work_on_vector(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,
+// Rust attaches additional meaning to the argument of `work_on_vector`: The function can assume
+// that it entirely *owns* `v`, and hence can do anything with it. When `work_on_vector` 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
+// Passing a `Vec<i32>` to `work_on_vector` 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!
+// If you give a book to your friend, you cannot just come 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
+// line in `ownership_demo`. 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.
+// ## Shared borrowing
// 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
// 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.
+// So, let's re-write `vec_min` to work on a shared borrow of a vector, written `&Vec<i32>`.
+// I also took the liberty to convert the function from `SomethingOrNothing` to the standard
+// library type `Option`.
fn vec_min(v: &Vec<i32>) -> Option<i32> {
let mut min = None;
for e in v {
+ // In the loop, `e` now has type `&i32`, so we have to dereference it to obtain an `i32`.
min = Some(match min {
None => *e,
Some(n) => cmp::min(n, *e)
}
// 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() {
+fn shared_borrow_demo() {
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.
+// What's going on here? First, `&` is how you create a shared borrow to something. All borrows are created like
+// this - there is no way to have something like a NULL pointer. 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 respective 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.
+// borrow, Rust knows that it cannot mutate `v` in any way. Hence the pointer into the buffer of `v`
+// that was created before calling `vec_min` remains valid.
+// ## Mutable borrowing
// 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.
+// 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.
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() {
+// Here's an example of calling `vec_inc`.
+fn mutable_borrow_demo() {
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.
+// `&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, and watch Rust complain. 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. In other words, Rust keeps us safe from bugs like the one in the C++ snipped above.
//
// 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);
-}
+// you may say: "Wait! Don't the `v` in `mutable_borrow_demo` and the `v` in `vec_inc` alias?" And you are right,
+// they do. However, the `v` in `mutable_borrow_demo` 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`.
-// So, to summarize: The ownership and borrowing system of Rust enforces the following three rules:
+// ## Summary
+// 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.
+// to tackle many problems beyond basic memory safety. You will see some examples for this soon.
-// [index](main.html) | [previous](part03.html) | [next](main.html)
+// [index](main.html) | [previous](part03.html) | [next](part05.html)
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