X-Git-Url: https://git.ralfj.de/rust-101.git/blobdiff_plain/767e5b9549d03685a601ccc2f32123f1dac618c6..46c141eefadadaf82b1414ae19d3766bbd4ba0cc:/src/part04.rs diff --git a/src/part04.rs b/src/part04.rs index a4bd164..e4c6a59 100644 --- a/src/part04.rs +++ b/src/part04.rs @@ -8,15 +8,14 @@ use std::cmp; // prevent bugs like this C++ snippet. /* void foo(std::vector 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` @@ -24,27 +23,29 @@ use std::cmp; // 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) { /* do something */ } -fn foo1() { +fn work_on_vector(v: Vec) { /* 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` to `take` is considered *transfer of ownership*: Someone used +// Passing a `Vec` 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 @@ -62,12 +63,13 @@ fn foo1() { // 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`. +// I also took the liberty to convert the function from `SomethingOrNothing` to the standard +// library type `Option`. fn vec_min(v: &Vec) -> Option { 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) @@ -77,7 +79,7 @@ fn vec_min(v: &Vec) -> Option { } // 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); @@ -87,61 +89,54 @@ fn foo2() { // 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. +// 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` is the type of mutable borrows of `vec`. Because the borrow is +// mutable, we can change `e` in the loop. fn vec_inc(v: &mut Vec) { for e in v { *e += 1; } } -// The type `&mut Vec` is the type of mutable borrows of `vec`. 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)