Merge pull request #3 from nelsonjchen/typo_fix

[rust-101.git] / src / part06.rs

diff --git a/src/part06.rs b/src/part06.rs
index abe4c631c2b85a570b9462f55e94b95770681c13..a4ac7641e68a6063bb943717af498e483fa96056 100644 (file)
--- a/src/part06.rs
+++ b/src/part06.rs
@@ -5,14 +5,14 @@
 use part05::BigInt;
 
 // With `BigInt` being about numbers, we should be able to write a version of `vec_min`
 use part05::BigInt;
 
 // With `BigInt` being about numbers, we should be able to write a version of `vec_min`

-// that computes the minimum of a list of `BigInt`. We start by writing `min` for
-// `BigInt`. Now our assumption of having no trailing zeros comes in handy!
+// that computes the minimum of a list of `BigInt`. First, we have to write `min` for `BigInt`.

 impl BigInt {
     fn min_try1(self, other: Self) -> Self {
 impl BigInt {
     fn min_try1(self, other: Self) -> Self {

-        // Just to be sure, we first check that both operands actually satisfy our invariant. `debug_assert!` is a
-        // macro that checks that its argument (must be of type `bool`) is `true`, and panics otherwise. It gets
-        // removed in release builds, which you do with `cargo build --release`.
+        //@ Just to be sure, we first check that both operands actually satisfy our invariant. `debug_assert!` is a
+        //@ macro that checks that its argument (must be of type `bool`) is `true`, and panics otherwise. It gets
+        //@ removed in release builds, which you do with `cargo build --release`.

         debug_assert!(self.test_invariant() && other.test_invariant());
         debug_assert!(self.test_invariant() && other.test_invariant());

+        // Now our assumption of having no trailing zeros comes in handy:

         // If the lengths of the two numbers differ, we already know which is larger.
         if self.data.len() < other.data.len() {
             self
         // If the lengths of the two numbers differ, we already know which is larger.
         if self.data.len() < other.data.len() {
             self


@@ -25,88 +25,88 @@ impl BigInt {
     }
 }
 
     }
 }
 

-// Now we can write `vec_min`. In order to make it type-check, we have to write it as follows.
+// Now we can write `vec_min`.

 fn vec_min(v: &Vec<BigInt>) -> Option<BigInt> {
     let mut min: Option<BigInt> = None;
 fn vec_min(v: &Vec<BigInt>) -> Option<BigInt> {
     let mut min: Option<BigInt> = None;

+    // If `v` is a shared borrowed vector, then the default for iterating over it is to call `iter`, the iterator that borrows the elements.

     for e in v {
     for e in v {

-        min = Some(match min {
-            None => e.clone(),
-            Some(n) => e.clone().min_try1(n)
-        });
+        let e = e.clone();
+        min = Some(match min {                                      /*@*/
+            None => e,                                              /*@*/
+            Some(n) => e.min_try1(n)                                /*@*/
+        });                                                         /*@*/

     }
     min
 }
     }
     min
 }

-// Now, what's happening here? Why do we have to write `clone()`, and why did we not
-// have to write that in our previous version?
-// 
-// The answer is already hidden in the type of `vec_min`: `v` is just borrowed, but
-// the Option<BigInt> that it returns is *owned*. We can't just return one of the elements of `v`,
-// as that would mean that it is no longer in the vector! In our code, this comes up when we update
-// the intermediate variable `min`, which also has type `Option<BigInt>`. If you replace `e.clone()`
-// in the `None` arm with `*e`, Rust will complain "Cannot move out of borrowed content". That's because
-// `e` is a `&BigInt`. Assigning `min = Some(*e)` works just like a function call: Ownership of the
-// underlying data is transferred from where `e` borrows from to `min`. But that's not allowed, since
-// we just borrowed `e`, so we cannot empty it! We can, however, call `clone()` on it. Then we own
-// the copy that was created, and hence we can store it in `min`.<br/>
-// Of course, making such a full copy is expensive, so we'd like to avoid it. We'll some to that in the next part.
+//@ Now, what's happening here? Why do we have to to make a full (deep) copy of `e`, and why did we not
+//@ have to do that in our previous version?
+//@ 
+//@ The answer is already hidden in the type of `vec_min`: `v` is just borrowed, but
+//@ the Option<BigInt> that it returns is *owned*. We can't just return one of the elements of `v`,
+//@ as that would mean that it is no longer in the vector! In our code, this comes up when we update
+//@ the intermediate variable `min`, which also has type `Option<BigInt>`. If you replace get rid of the
+//@ `e.clone()`, Rust will complain "Cannot move out of borrowed content". That's because
+//@ `e` is a `&BigInt`. Assigning `min = Some(*e)` works just like a function call: Ownership of the
+//@ underlying data is transferred from where `e` borrows from to `min`. But that's not allowed, since
+//@ we just borrowed `e`, so we cannot empty it! We can, however, call `clone` on it. Then we own
+//@ the copy that was created, and hence we can store it in `min`. <br/>
+//@ Of course, making such a full copy is expensive, so we'd like to avoid it. We'll come to that in the next part.

 
 // ## `Copy` types
 
 // ## `Copy` types

-// But before we go there, I should answer the second question I brought up above: Why did our old `vec_min` work?
-// We stored the minimal `i32` locally without cloning, and Rust did not complain. That's because there isn't
-// really much of an "ownership" when it comes to types like `i32` or `bool`: If you move the value from one
-// place to another, then both instance are "complete". We also say the value has been *duplicated*. This is in
-// stark contrast to types like `Vec<i32>`, where moving the value results in both the old and the new vector to
-// point to the same underlying buffer. We don't have two vectors, there's no duplication.
-//
-// Rust calls types that can be freely duplicated `Copy` types. `Copy` is another trait, and it is implemented for
-// types like `i32` and `bool`. Remember how we defined the trait `Minimum` by writing `trait Minimum : Copy { ...`?
-// This tells Rust that every type that implements `Minimum` must also implement `Copy`, and that's why the compiler
-// accepted our generic `vec_min` in part 02. `Copy` is the first *marker trait* that we encounter: It does not provide
-// any methods, but makes a promise about the behavior of the type - in this case, being duplicable.
+//@ But before we go there, I should answer the second question I brought up above: Why did our old `vec_min` work?
+//@ We stored the minimal `i32` locally without cloning, and Rust did not complain. That's because there isn't
+//@ really much of an "ownership" when it comes to types like `i32` or `bool`: If you move the value from one
+//@ place to another, then both instances are "complete". We also say the value has been *duplicated*. This is in
+//@ stark contrast to types like `Vec<i32>`, where moving the value results in both the old and the new vector to
+//@ point to the same underlying buffer. We don't have two vectors, there's no proper duplication.
+//@
+//@ Rust calls types that can be easily duplicated `Copy` types. `Copy` is another trait, and it is implemented for
+//@ types like `i32` and `bool`. Remember how we defined the trait `Minimum` by writing `trait Minimum : Copy { ...`?
+//@ This tells Rust that every type that implements `Minimum` must also implement `Copy`, and that's why the compiler
+//@ accepted our generic `vec_min` in part 02. `Copy` is the first *marker trait* that we encounter: It does not provide
+//@ any methods, but makes a promise about the behavior of the type - in this case, being duplicable.

 
 

-// If you try to implement `Copy` for `BigInt`, you will notice that Rust
-// does not let you do that. A type can only be `Copy` if all its elements
-// are `Copy`, and that's not the case for `BigInt`. However, we can make
-// `SomethingOrNothing<T>` copy if `T` is `Copy`.
+//@ If you try to implement `Copy` for `BigInt`, you will notice that Rust
+//@ does not let you do that. A type can only be `Copy` if all its elements
+//@ are `Copy`, and that's not the case for `BigInt`. However, we can make
+//@ `SomethingOrNothing<T>` copy if `T` is `Copy`.

 use part02::{SomethingOrNothing,Something,Nothing};
 use part02::{SomethingOrNothing,Something,Nothing};

-impl<T: Copy> Copy for SomethingOrNothing<T>{}
-// Again, Rust can generate implementations of `Copy` automatically. If
-// you add `#[derive(Copy,Clone)]` right before the definition of `SomethingOrNothing`,
-// both `Copy` and `Clone` will automatically be implemented.
+impl<T: Copy> Copy for SomethingOrNothing<T> {}
+//@ Again, Rust can generate implementations of `Copy` automatically. If
+//@ you add `#[derive(Copy,Clone)]` right before the definition of `SomethingOrNothing`,
+//@ both `Copy` and `Clone` will automatically be implemented.

 
 

-// ## An operational perspective
-// Instead of looking at what happens "at the surface" (i.e., visible in Rust), one can also explain
-// ownership passing and how `Copy` and `Clone` fit by looking at what happens on the machine.<br/>
-// When Rust code is executed, passing a value (like `i32` or `Vec<i32>`) to a function will always
-// result in a shallow copy being performed: Rust just copies the bytes representing that value, and
-// considers itself done. That's just like the default copy constructor in C++. Rust, however, will
-// consider this a destructive operation: After copying the bytes elsewhere, the original value must
-// no longer be used. After all, the two could not share a pointer! If, however, you mark a type `Copy`,
-// then Rust will *not* consider a move destructive, and just like in C++, the old and new value
-// can happily coexist. Now, Rust does not allow to to overload the copy constructor. This means that
-// passing a value around will always be a fast operation, no allocation or any other kind of heap access
-// will happen. In the situations where you would write a copy constructor in C++ (and hence
-// incur a hidden cost on every copy of this type), you'd have the type *not* implement `Copy`, but only
-// `Clone`. This makes the cost explicit.
+//@ ## An operational perspective
+//@ Instead of looking at what happens "at the surface" (i.e., visible in Rust), one can also explain
+//@ ownership passing and how `Copy` and `Clone` fit in by looking at what happens on the machine. <br/>
+//@ When Rust code is executed, passing a value (like `i32` or `Vec<i32>`) to a function will always
+//@ result in a shallow copy being performed: Rust just copies the bytes representing that value, and
+//@ considers itself done. That's just like the default copy constructor in C++. Rust, however, will
+//@ consider this a destructive operation: After copying the bytes elsewhere, the original value must
+//@ no longer be used. After all, the two could now share a pointer! If, however, you mark a type `Copy`,
+//@ then Rust will *not* consider a move destructive, and just like in C++, the old and new value
+//@ can happily coexist. Now, Rust does not allow you to overload the copy constructor. This means that
+//@ passing a value around will always be a fast operation, no allocation or any other kind of heap access
+//@ will happen. In the situations where you would write a copy constructor in C++ (and hence
+//@ incur a hidden cost on every copy of this type), you'd have the type *not* implement `Copy`, but only
+//@ `Clone`. This makes the cost explicit.

 
 // ## Lifetimes
 
 // ## Lifetimes

-// To fix the performance problems of `vec_min`, we need to avoid using `clone()`. We'd like
-// the return value to not be owned (remember that this was the source of our need for cloning), but *borrowed*.
+//@ To fix the performance problems of `vec_min`, we need to avoid using `clone`. We'd like
+//@ the return value to not be owned (remember that this was the source of our need for cloning), but *borrowed*.

 
 

-// This is demonstrated by the function `head` that borrows the first element of a vector if it is non-empty.
-// The type of the function says that it will either return nothing, or it will return a borrowed `T`.
-// We can then borrow the first element of `v` and use it to construct the return value.
+//@ The function `head` demonstrates how that could work: It borrows the first element of a vector if it is non-empty.
+//@ The type of the function says that it will either return nothing, or it will return a borrowed `T`.
+//@ We can then borrow the first element of `v` and use it to construct the return value.

 fn head<T>(v: &Vec<T>) -> Option<&T> {
     if v.len() > 0 {
 fn head<T>(v: &Vec<T>) -> Option<&T> {
     if v.len() > 0 {

-        Some(&v[0])
+        Some(&v[0])                                                 /*@*/

     } else {
         None
     }
 }
     } else {
         None
     }
 }

-
-// Now, coming back to `head` - here, we are returning a pointer to the first element. But doesn't
-// that mean that callers have to be careful? Imagine `head` would be a C++ function, and we would
-// write the following code.
+// Technically, we are returning a pointer to the first element. But doesn't that mean that callers have to be
+// careful? Imagine `head` would be a C++ function, and we would write the following code.

 /*
   int foo(std::vector<int> v) {
     int *first = head(v);
 /*
   int foo(std::vector<int> v) {
     int *first = head(v);


@@ -114,36 +114,37 @@ fn head<T>(v: &Vec<T>) -> Option<&T> {
     return *first;
   }
 */
     return *first;
   }
 */

-// This is very much like our very first motivating example for ownership, at the beginning of part 04.
-// But this time, the bug is hidden behind the call to `head`. How does Rust solve this? If we translate
-// the code above to Rust, it doesn't compile, so clearly we are good - but how and why?
-// (Notice that have to explicitly assert using `unwrap` that `first` is not `None`, whereas the C++ code
-// above would silently dereference a `NULL`-pointer. But that's another point.)
+//@ This is very much like our very first motivating example for ownership, at the beginning of part 04:
+//@ `push_back` could reallocate the buffer, making `first` an invalid pointer. Again, we have aliasing (of `first`
+//@ and `v`) and mutation. But this time, the bug is hidden behind the call to `head`. How does Rust solve this? If we translate
+//@ the code above to Rust, it doesn't compile, so clearly we are good - but how and why?
+//@ (Notice that have to explicitly assert using `unwrap` that `first` is not `None`, whereas the C++ code
+//@ above would silently dereference a `NULL`-pointer. But that's another point.)

 fn rust_foo(mut v: Vec<i32>) -> i32 {
     let first: Option<&i32> = head(&v);
     /* v.push(42); */
     *first.unwrap()
 }
 
 fn rust_foo(mut v: Vec<i32>) -> i32 {
     let first: Option<&i32> = head(&v);
     /* v.push(42); */
     *first.unwrap()
 }
 

-// To give the answer to this question, we have to talk about the *lifetime* of a borrow. The point is, saying that
-// you borrowed your friend a `Vec<i32>`, or a book, is not good enough, unless you also agree on *how long*
-// your friend can borrow. After all, you need to know when you can rely on owning your data (or book) again.
-// 
-// Every borrow in Rust has an associated lifetime. The full type of `head` reads as follows:
-// `fn<'a, T>(&'a Vec<T>) -> Option<&'a T>`. Here, `'a` is a *lifetime variable*, which represents how long the vector has
-// been borrowed. The function type expresses that argument and return value have *the same lifetime*.
-// 
-// When analyzing the code of `rust_foo`, Rust has to assign a lifetime to `first`. It will choose the scope
-// where `first` is valid, which is the entire rest of the function. Because `head` ties the lifetime of its
-// argument and return value together, this means that `&v` also has to borrow `v` for the entire duration of
-// the function. So when we try to borrow `v` mutable for `push`, Rust complains that the two borrows (the one
-// for `head`, and the one for `push`) overlap. Lucky us! Rust caught our mistake and made sure we don't crash the program.
-// 
-// So, to sum this up: Lifetimes enable Rust to reason about *how long* a pointer has been borrowed. We can thus
-// safely write functions like `head`, that return pointers into data they got as argument, and make sure they
-// are used correctly, *while looking only at the function type*. At no point in our analysis of `rust_foo` did
-// we have to look *into* `head`. That's, of course, crucial if we want to separate library code from application code.
-// Most of the time, we don't have to explicitly add lifetimes to function types. This is thanks to *lifetimes elision*,
-// where Rust will automatically insert lifetimes we did not specify, following some [simple, well-documented rules](http://doc.rust-lang.org/stable/book/lifetimes.html#lifetime-elision).
+//@ To give the answer to this question, we have to talk about the *lifetime* of a borrow. The point is, saying that
+//@ you borrowed your friend a `Vec<i32>`, or a book, is not good enough, unless you also agree on *how long*
+//@ your friend can borrow it. After all, you need to know when you can rely on owning your data (or book) again.
+//@ 
+//@ Every borrow in Rust has an associated lifetime, written `&'a T` for a borrow of type `T` with lifetime `'a`. The full
+//@ type of `head` reads as follows: `fn<'a, T>(&'a Vec<T>) -> Option<&'a T>`. Here, `'a` is a *lifetime variable*, which
+//@ represents how long the vector has been borrowed. The function type expresses that argument and return value have *the same lifetime*.
+//@ 
+//@ When analyzing the code of `rust_foo`, Rust has to assign a lifetime to `first`. It will choose the scope
+//@ where `first` is valid, which is the entire rest of the function. Because `head` ties the lifetime of its
+//@ argument and return value together, this means that `&v` also has to borrow `v` for the entire duration of
+//@ the function. So when we try to borrow `v` mutable for `push`, Rust complains that the two borrows (the one
+//@ for `head`, and the one for `push`) overlap. Lucky us! Rust caught our mistake and made sure we don't crash the program.
+//@ 
+//@ So, to sum this up: Lifetimes enable Rust to reason about *how long* a pointer has been borrowed. We can thus
+//@ safely write functions like `head`, that return pointers into data they got as argument, and make sure they
+//@ are used correctly, *while looking only at the function type*. At no point in our analysis of `rust_foo` did
+//@ we have to look *into* `head`. That's, of course, crucial if we want to separate library code from application code.
+//@ Most of the time, we don't have to explicitly add lifetimes to function types. This is thanks to *lifetimes elision*,
+//@ where Rust will automatically insert lifetimes we did not specify, following some [simple, well-documented rules](https://doc.rust-lang.org/stable/book/lifetimes.html#lifetime-elision).

 
 

-// [index](main.html) | [previous](part05.html) | [next](part07.html)
+//@ [index](main.html) | [previous](part05.html) | [raw source](https://www.ralfj.de/git/rust-101.git/blob_plain/HEAD:/workspace/src/part06.rs) | [next](part07.html)