add trim() to reading from stdin, so that it also works on Windows

[rust-101.git] / src / part04.rs

diff --git a/src/part04.rs b/src/part04.rs
index 47177a7731d27172fd2d448f364415553fcfe7f9..f39df37148f89efd586bc0af778226272929ae0e 100644 (file)
--- a/src/part04.rs
+++ b/src/part04.rs
@@ -1,11 +1,9 @@
 // Rust-101, Part 04: Ownership, Borrowing
 // =======================================
 
 // 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.
+//@ 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];
 /*
   void foo(std::vector<int> v) {
       int *first = &v[0];


@@ -13,63 +11,65 @@ use std::cmp;
       *first = 1337; // This is bad!
   }
 */
       *first = 1337; // This is bad!
   }
 */

-// 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`
-// 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'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`
+//@ 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.
+// ## Ownership
+//@ What does that mean in practice? Consider the following example.

 fn work_on_vector(v: Vec<i32>) { /* do something */ }
 fn ownership_demo() {
     let v = vec![1,2,3,4];
     work_on_vector(v);
     /* println!("The first element is: {}", v[0]); */
 }
 fn work_on_vector(v: Vec<i32>) { /* do something */ }
 fn ownership_demo() {
     let v = vec![1,2,3,4];
     work_on_vector(v);
     /* println!("The first element is: {}", v[0]); */
 }

-// 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 `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 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 `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.
+//@ 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 `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 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 `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.

 
 

-// 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.
+// ## 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
+//@ 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. `&Vec<i32>` says that we need
-// a vector, but we won't own it. I also took the liberty to convert the function from
-// `SomethingOrNothing` to the standard library type `Option`.
+//@ 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> {
 fn vec_min(v: &Vec<i32>) -> Option<i32> {

+    use std::cmp;
+

     let mut min = None;
     for e in v {
     let mut min = None;
     for e in v {

-        // In the loop, `e` now has type `&i32`, so we have to dereference it.
+        // 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)
         min = Some(match min {
             None => *e,
             Some(n) => cmp::min(n, *e)


@@ -86,27 +86,29 @@ fn shared_borrow_demo() {
     vec_min(&v);
     println!("The first element is: {}", *first);
 }
     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
-// 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 into the buffer of `v`
-// that was created before calling `vec_min` remains valid.
+//@ 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 into the buffer of `v`
+//@ that was 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.
+// ## 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.
+//@ 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;
     }
 }
 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?
+// 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]; */
 fn mutable_borrow_demo() {
     let mut v = vec![5,4,3,2,1];
     /* let first = &v[0]; */


@@ -114,33 +116,21 @@ fn mutable_borrow_demo() {
     vec_inc(&mut v);
     /* println!("The first element is: {}", *first); */
 }
     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, 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.
-// 
-// 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 `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`. This is, in fact, what
-// prevents the creation of a mutable borrow when there already is a shared one, as you witnessed
-// when enabling `first` above.
-
-// This also works the other way around: In `multiple_borrow_demo`, 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 multiple_borrow_demo() {
-    let mut v = vec![5,4,3,2,1];
-    let first = &mut v[0];
-    /* vec_min(&v); */
-    *first += 1;
-    println!("The first element is now: {}", *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, 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 `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
 // 
 // * 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


@@ -149,4 +139,4 @@ fn multiple_borrow_demo() {
 // 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. You will see some examples for this soon.
 
 // 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. 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)