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Can I destructure a tuple without binding the result to a new variable in a let/match/for statement?
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I’m getting an error I’m not sure how to handle, related to a tuple assignment I’m trying. The incr function below gets a left-hand of expression not valid error. What am I misunderstanding?
struct Fib {
i: u64,
fa: u64,
fb: u64,
}
impl Fib {
fn incr(&mut self) {
self.i += 1;
(self.fa, self.fb) = (self.fa + self.fb, self.fa);
}
}
As the helpful error explanation says†, you try to assign to a non-place expression. A place expression represents a memory location, and thus it can be a variable, a dereference, an indexing expression or a field reference, but a tuple is not one of these.
If you would use a binding, such as:
let (x, y) = (1, 2);
that would be a whole different story because let statements have different rules than assignment: the left hand side of a let statement is a pattern, not an expression, and (x, y) is a legal pattern.
To solve your problem, you may want to do the following and introduce a temporary variable, and then update the values of the members:
(The following is also fixing your fibonacci sequence, i.e. correcting the values of the members since they are naturally ordered as 'a' and 'b')
impl Fib {
fn incr(&mut self) {
self.i += 1;
let fa = self.fa;
self.fa = self.fb;
self.fb += fa;
}
}
Note: Albeit it was not your question, I would strongly advise to implement Iterator for your Fib type, in which case you wouldn't have to keep track of the index (i) because that would be available through the enumerate method.
E.g.
impl Iterator for Fib {
type Item = u64;
fn next(&mut self) -> Option<Self::Item> {
let fa = self.fa;
self.fa = self.fb;
self.fb += fa;
Some(fa)
}
}
And then you could use it as:
for (i, x) in my_fib.enumerate() { ... }
† rustc --explain E0070
Related
So I have this enum that defines different view positions on a View controller when a side bar menu is presented. I need to add, subtract, multiply, or divide the different values based on different situations. How exactly do I form a method to allow me to use -, +, *, or / operators on the values in the enum. I can find plenty examples that use the compare operator ==. Although I haven't been able to find any that use >=. Which I also need to be able to do.
Here is the enum
enum FrontViewPosition: Int {
case None
case LeftSideMostRemoved
case LeftSideMost
case LeftSide
case Left
case Right
case RightMost
case RightMostRemoved
}
Now I'm trying to use these operators in functions like so.
func getAdjustedFrontViewPosition(_ frontViewPosition: FrontViewPosition, forSymetry symetry: Int) {
var frontViewPosition = frontViewPosition
if symetry < 0 {
frontViewPosition = .Left + symetry * (frontViewPosition - .Left)
}
}
Also in another function like so.
func rightRevealToggle(animated: Bool) {
var toggledFrontViewPosition: FrontViewPosition = .Left
if self.frontViewPosition >= .Left {
toggledFrontViewPosition = .LeftSide
}
self.setFrontViewPosition(toggledFrontViewPosition, animated: animated)
}
I know that i need to directly create the functions to allow me to use these operators. I just don't understand how to go about doing it. A little help would be greatly appreciated.
The type you are trying to define has a similar algebra to pointers in that you can add an offset to a pointer to get a pointer and subtract two pointers to get a difference. Define these two operators on your enum and your other functions will work.
Any operators over your type should produce results in your type. There are different ways to achieve this, depending on your requirements. Here we shall treat your type as a wrap-around ("modulo") one - add 1 to the last literal and you get the first. To do this we use raw values from 0 to n for your types literals and use modulo arithmetic.
First we need a modulo operator which always returns a +ve result, the Swift % can return a -ve one which is not what is required for modulo arithmetic.
infix operator %% : MultiplicationPrecedence
func %%(_ a: Int, _ n: Int) -> Int
{
precondition(n > 0, "modulus must be positive")
let r = a % n
return r >= 0 ? r : r + n
}
Now your enum assigning suitable raw values:
enum FrontViewPosition: Int
{
case None = 0
case LeftSideMostRemoved = 1
case LeftSideMost = 2
case LeftSide = 3
case Left = 4
case Right = 5
case RightMost = 6
case RightMostRemoved = 7
Now we define the appropriate operators.
For addition we can add an integer to a FrontViewPosition and get a FrontViewPosition back. To do this we convert to raw values, add, and then reduce modulo 8 to wrap-around. Note the need for a ! to return a non-optional FrontViewPosition - this will always succeed due to the modulo math:
static func +(_ x : FrontViewPosition, _ y : Int) -> FrontViewPosition
{
return FrontViewPosition(rawValue: (x.rawValue + y) %% 8)!
}
For subtraction we return the integer difference between two FrontViewPosition values:
static func -(_ x : FrontViewPosition, _ y : FrontViewPosition) -> Int
{
return x.rawValue - y.rawValue
}
}
You can define further operators as needed, say a subtraction operator which takes a FrontViewPosition and an Int and returns a FrontViewPosition.
HTH
Enum could have function~
enum Tst:Int {
case A = 10
case B = 20
case C = 30
static func + (t1:Tst,t2:Tst) -> Tst {
return Tst.init(rawValue: t1.rawValue+t2.rawValue)! //here could be wrong!
}
}
var a = Tst.A
var b = Tst.B
var c = a+b
I'm new to Rust and looking to understand concepts like borrowing. I'm trying to create a simple two dimensional array using standard input. The code:
use std::io;
fn main() {
let mut values = [["0"; 6]; 6]; // 6 * 6 array
// iterate 6 times for user input
for i in 0..6 {
let mut outputs = String::new();
io::stdin().read_line(&mut outputs).expect(
"failed to read line",
);
// read space separated list 6 numbers. Eg: 5 7 8 4 3 9
let values_itr = outputs.trim().split(' ');
let mut j = 0;
for (_, value) in values_itr.enumerate() {
values[i][j] = value;
j += 1;
}
}
}
This won't compile because the outputs variable lifetime is not long enough:
error[E0597]: `outputs` does not live long enough
--> src/main.rs:20:5
|
14 | let values_itr = outputs.trim().split(' ');
| ------- borrow occurs here
...
20 | }
| ^ `outputs` dropped here while still borrowed
21 | }
| - borrowed value needs to live until here
How can I get the iterated values out of the block into values array?
split() gives you substrings (string slices) borrowed from the original string, and the original string is outputs from line 6.
The string slices can't outlive the scope of outputs: when a loop iteration ends, outputs is deallocated.
Since values is longer lived, the slices can't be stored there.
We can't borrow slices of outputs across a modification of outputs. So even if the String outputs itself was defined before values, we couldn't easily put the string slices from .split() into values; modifying the string (reading into it) invalidates the slices.
A solution needs to either
Use a nested array of String, and when you assign an element from the split iterator, make a String from the &str using .to_string(). I would recommend this solution. (However an array of String is not at as easy to work with, maybe already this requires using Vec instead.) 1
Read all input before constructing a nested array of &str that borrows from the input String. This is good if the nested array is something that you only need temporarily.
1: You can use something like vec![vec![String::new(); 6]; 6] instead
This answer was moved from the question, where it solved the OPs needs.
use std::io;
fn main() {
let mut values = vec![vec![String::new(); 6]; 6];
for i in 0..6 {
let mut outputs = String::new();
io::stdin().read_line(&mut outputs)
.expect("failed to read line");
let values_itr = outputs.trim().split(' ');
let mut j = 0;
for (_, value) in values_itr.enumerate() {
values[i][j] = value.to_string();
j += 1;
}
}
}
I have just started to learn Rust. During my first steps with this language, I found a strange behaviour, when an iteration is performed inside main or in another function as in following example:
fn myfunc(x: &Vec<f64>) {
let n = x.len();
println!(" n: {:?}", n);
for i in -1 .. n {
println!(" i: {}", i);
}
}
fn main() {
for j in -1 .. 6 {
println!("j: {}", j);
}
let field = vec![1.; 6];
myfunc(&field);
}
While the loop in main is correctly displayed, nothing is printed for the loop inside myfunc and I get following output:
j: -1
j: 0
j: 1
j: 2
j: 3
j: 4
j: 5
n: 6
What is the cause of this behaviour?
Type inference is causing both of the numbers in your range to be usize, which cannot represent negative numbers. Thus, the range is from usize::MAX to n, which never has any members.
To find this out, I used a trick to print out the types of things:
let () = -1 .. x.len();
Which has this error:
error: mismatched types:
expected `core::ops::Range<usize>`,
found `()`
(expected struct `core::ops::Range`,
found ()) [E0308]
let () = -1 .. x.len();
^~
Diving into the details, slice::len returns a usize. Your -1 is an untyped integral value, which will conform to fit whatever context it needs (if there's nothing for it to conform to, it will fall back to an i32).
In this case, it's as if you actually typed (-1 as usize)..x.len().
The good news is that you probably don't want to start at -1 anyway. Slices are zero-indexed:
fn myfunc(x: &[f64]) {
let n = x.len();
println!(" n: {:?}", n);
for i in 0..n {
println!(" i: {}", i);
}
}
Extra good news is that this annoyance was fixed in the newest versions of Rust. It will cause a warning and then eventually an error:
warning: unary negation of unsigned integers will be feature gated in the future
for i in -1 .. n {
^~
Also note that you should never accept a &Vec<T> as a parameter. Always use a &[T] as it's more flexible and you lose nothing.
I'm coming from a C (and to a lesser extent, C++) background. I wrote the following code snippet:
fn main() {
let my_array = [1, 2, 3];
let print_me = |j| println!("= {}", j);
for k in my_array.iter() {
print_me(k);
}
}
This compiled and ran as expected, but then I specified the type of the argument passed to the closure print_me thus:
fn main() {
let my_array = [1, 2, 3];
let print_me = |j: i32| println!("= {}", j);
for k in my_array.iter() {
print_me(k);
}
}
I got a compilation error:
error[E0308]: mismatched types
--> src/main.rs:6:22
|
6 | print_me(k);
| ^
| |
| expected i32, found &{integer}
| help: consider dereferencing the borrow: `*k`
|
= note: expected type `i32`
found type `&{integer}`
Now this confused me until I changed k to &k in the for statement, which worked fine:
fn main() {
let my_array = [1, 2, 3];
let print_me = |j: i32| println!("= {}", j);
for &k in my_array.iter() {
print_me(k);
}
}
It seems that I misunderstood the for syntax itself -- or maybe the exact workings of an iterator -- or maybe the usage syntax of a reference vis-a-vis a pointer [which are related but distinct in C++].
In the construct for A in B { C1; C2; ... Cn }, what exactly are A and B supposed to be?
First of all, here's a link to the definition of for in the reference.
To summarise, B is any expression which evaluates to something that can be converted into a value that implements the Iterator<T> trait, whilst A is a irrefutable pattern that binds values of type T.
In your specific case, slice::iter returns an Iter<i32>, which implements Iterator<Item = &i32>. That is, it doesn't yield i32s, it yields &i32s.
Thus, in both the first and second examples, k is actually binding to &i32s, not i32s. When you specified the type of the closure, you were actually specifying the wrong type. The reason the final example works is because A is a pattern, not a variable name. What &k is actually doing is "de-structuring" the &i32, binding the i32 part to a variable named k.
The "irrefutable" part simply means that the pattern must always work. For example, you can't do for Some(x) in thingy where thingy implements Iterator<Option<_>>; Some(x) would not necessarily be valid for every element in the iterator; thus, it's a refutable pattern.
Many iterators actually return a reference rather than a value. To be sure, you have to check the return type of .iter(), which should be of the form Iterator<Item = X>: X will be the type of the variable returned.
So here:
fn main() {
let my_array = [1, 2, 3];
let print_me = |j: i32| println!("= {}", j);
for k in my_array.iter() {
print_me(k);
}
}
This X is &i32 (a reference to i32), and therefore k has type &i32.
This is why, when calling print_me, there is an error: &i32 is passed where i32 is expected.
There are multiple possible fixes here:
specify a different type to print_me:
let print_me = |j: &i32| println!("= {}", j);
dereference the value of k:
print_me(*k);
change the type of k by destructuring in the loop:
for &k in my_array.iter() { ... }
The destructuring occurs because for .. in accepts an irrefutable pattern, so you can pattern match like you would do in a match expression, except that the variable's type has to match (otherwise you get a compiler time error).
To better illustrate it, we can use a slightly more complicated example:
fn main() {
let my_array = [(1, 2), (2, 3), (3, 4)];
let print_me = |a: i32, b: i32| println!("= {} {}", a, b);
for &(j, k) in my_array.iter() {
print_me(j, k)
}
}
The type of my_array is [(i32, i32)]: an array of tuples of 2 i32. The result of .iter() is therefore of type Iterator<Item = &(i32, i32)>: an iterator to a reference to a tuple of 2 i32 aka &(i32, i32).
When we use the irrefutable pattern &(j, k) what happens is that we destructure the tuple so that:
the first element binds to j (inferred to be of type i32, only works because i32 is Copy)
the second element binds to k ((inferred to be of type i32)
j and k thus become temporary copies of the i32 inside this element.
The Rust Guide states that:
The semicolon turns any expression into a statement by throwing away its value and returning unit instead.
I thought I got this concept down until I ran an experiment:
fn print_number(x: i32, y: i32) -> i32 {
if x + y > 20 { return x }
x + y
}
Which compiles fine. Then, I added a semicolon at the end of the return line (return x;). From what I understand, this turns the line into a statement, returning the unit data type ().
Nonetheless, the end result is the same.
Normally, every branch in the if expression should have the same type. If the type for some branch is underspecified, the compiler tries to find the single common type:
fn print_number(x: int, y: int) {
let v = if x + y > 20 {
3 // this can be either 3u, 3i, 3u8 etc.
} else {
x + y // this is always int
};
println!("{}", v);
}
In this code, 3 is underspecified but the else branch forces it to have the type of int.
This sounds simple: There is a function that "unifies" two or more types into the common type, or it will give you an error when that's not possible. But what if there were a fail! in the branch?
fn print_number(x: int, y: int) {
let v = if x + y > 20 {
fail!("x + y too large") // ???
} else {
x + y // this is always int
};
println!("{}", v); // uh wait, what's the type of `v`?
}
I'd want that fail! does not affect other branches, it is an exceptional case after all. Since this pattern is quite common in Rust, the concept of diverging type has been introduced. There is no value which type is diverging. (It is also called an "uninhabited type" or "void type" depending on the context. Not to be confused with the "unit type" which has a single value of ().) Since the diverging type is naturally a subset of any other types, the compiler conclude that v's type is just that of the else branch, int.
Return expression is no different from fail! for the purpose of type checking. It abruptly escapes from the current flow of execution just like fail! (but does not terminate the task, thankfully). Still, the diverging type does not propagate to the next statement:
fn print_number(x: int, y: int) {
let v = if x + y > 20 {
return; // this is diverging
() // this is implied, even when you omit it
} else {
x + y // this is always int
};
println!("{}", v); // again, what's the type of `v`?
}
Note that the sole semicoloned statement x; is equivalent to the expression x; (). Normally a; b has the same type as b, so it would be quite strange that x; () has a type of () only when x is not diverging, and it diverges when x does diverge. That's why your original code didn't work.
It is tempting to add a special case like that:
Why don't you make x; () diverging when x diverges?
Why don't you assume uint for every underspecified integer literal when its type cannot be inferred? (Note: this was the case in the past.)
Why don't you automatically find the common supertrait when unifying multiple trait objects?
The truth is that, designing the type system is not very hard, but verifying it is much harder and we want to ensure that Rust's type system is future-proof and long standing. Some of them may happen if it really is useful and it is proved "correct" for our purpose, but not immediately.
I'm not 100% sure of what I'm saying but it kinda makes sense.
There's an other concept coming into play: reachability analysis. The compiler knows that what follows a return expression statement is unreachable. For example, if we compile this function:
fn test() -> i32 {
return 1;
2
}
We get the following warning:
warning: unreachable expression
--> src/main.rs:3:5
|
3 | 2
| ^
|
The compiler can ignore the "true" branch of the if expression if it ends with a return expression and only consider the "false" branch when determining the type of the if expression.
You can also see this behavior with diverging functions. Diverging functions are functions that don't return normally (e.g. they always fail). Try replacing the return expression with the fail! macro (which expands to a call to a diverging function). In fact, return expressions are also considered to be diverging; this is the basis of the aforementioned reachability analysis.
However, if there's an actual () expression after the return statement, you'll get an error. This function:
fn print_number(x: i32, y: i32) -> i32 {
if x + y > 20 {
return x;
()
} else {
x + y
}
}
gives the following error:
error[E0308]: mismatched types
--> src/main.rs:4:9
|
4 | ()
| ^^ expected i32, found ()
|
= note: expected type `i32`
found type `()`
In the end, it seems diverging expressions (which includes return expressions) are handled differently by the compiler when they are followed by a semicolon: the statement is still diverging.