Im getting error in if(Liste.length>1), i want to make a function to return true if list is ordered low to high. Having some problems with Boolean returning.
object Soru5 extends App {
//A function returns true if list is ordered low to high
def kucukten_buyuge_siralimi(Liste:List[Int]):Boolean=
{
var i=0
var ListeIslem=Liste
if(Liste.length==0) false
if(Liste.length==1) true
if(Liste.length>1) //***********ERROR************ found->Unit,Boolean required
{
while(i<Liste.length-1)
{
var onceki=ListeIslem.head //onceki is first element of list
var sonraki=(ListeIslem.tail).head //sonraki is second element of list
if(onceki>sonraki)
{
return false
}
else
{
ListeIslem=ListeIslem.tail
i=i+1
}
}
return true
}
}
val listem=List(3,2,1)
println(kucukten_buyuge_siralimi(listem))
}
The return value of a method is the value of the last expression evaluated inside the method body. In your case, the last expression inside the method body is:
if(Liste.length>1) {
// …
}
So … what is the value of this expression if the length is less than 2? There is no value! Actually, in Scala, there is (almost) always a value, so the value of this is () which is of type Unit and denotes the absence of a value.
In other words, if the length is less than 2, your method will return () of type Unit, but it promises to return something of type Boolean. You need to add an else clause to your if expression that also returns a Boolean.
This is the immediate cause of the error.
Note that there are also some other problems with your code besides that. For example, those two lines don't actually do anything:
if(Liste.length==0) false
if(Liste.length==1) true
They simply evaluate to a value, but you don't actually do anything with that value, you don't store it in a variable, you don't return it, you don't pass it as an argument … it just vanishes.
Note also that your code violates the Community Coding Style.
And last but not least, your code just doesn't read like Scala. It reads more like a mix of Fortran and Basic. Scala style would be more like this:
def isMonotonicallyIncreasing(l: List[Int]) =
if (l.size < 2) true else l.sliding(2).forall { case (a :: b :: Nil) ⇒ a <= b }
i.e. a list is monotonically increasing if for all subsequent pairs of elements, the left element is not greater than the right element.
Because there's no else between ifs, they are separate. You effectively have
if (Liste.length==0) false else ();
if (Liste.length==1) true else ();
if (Liste.length>1) ... else ();
else () is what the compiler inserts for every if without else.
The compiler has no idea that Liste.length will be the same each time or that it will never be negative. It would be better to write
Liste.length match {
case 0 => false
case 1 => true
case n => ...
You can use n in the last branch instead of calculating length yet again.
This can be improved further, of course, but I think it's out of scope for this answer.
Related
What is the mechanism of declaring w/o value in Swift5 ? Does the first assign become the real declaration ?
And, should we avoid to declare without value in Swift?
var v:String;
if true {
v = "Hello"
print(v) // print "Hello" when without the print below
}
print(v) // variable 'v' used before being initialized
var v:String="";
if true {
v = "Hello"
print(v) // print "Hello"
}
print(v) // print "Hello"
Well, the message is not very helpful, and that's the problem. This pattern (which I call computed initialization) is perfectly legal and useful and — amazingly — you can even use let instead of var. But you must initialize the uninitialized variable by all possible paths before you use it. So you have:
var v:String
if true {
v = "Hello"
}
print(v) // error
But hold my beer and watch this:
var v:String
if true {
v = "Hello"
} else {
v = "Goodbye"
}
print(v) // fine!
Or even:
let v:String
if true {
v = "Hello"
} else {
v = "Goodbye"
}
print(v) // fine!
Amazing, eh?
Now, you might say: OK, but true will always be true so it's silly to make me fulfill the "all paths" rule. Too bad! The compiler insists anyway, and then lets you off later with a warning that the else won't be executed. But a warning lets you compile; an error doesn't. The truth is that your example is very artificial. But this is a real-life possibility:
let v:String
if self.someBoolProperty {
v = "Hello"
} else {
v = "Goodbye"
}
print(v) // fine!
Not only is this sort of thing legal, it is actually the pattern that Apple recommends under certain slightly tricky circumstances. For instance, it is used in Apple's own example code showing how to use the Swift 5 Result struct:
let result: Result<Int, EntropyError>
if count < AsyncRandomGenerator.entropyLimit {
// Produce numbers until reaching the entropy limit.
result = .success(Int.random(in: 1...100))
} else {
// Supply a failure reason when the caller hits the limit.
result = .failure(.entropyDepleted)
}
So this is because swift compiles your code and notices that your value var v:String; is undeclared before being used which makes it an "optional" value. Even though you are assigning it within the if statement, if you were to get rid of the true value it is possible that the if statement would never run therefore no value will ever be stored in v, thus it would be used before "assigned".
So to answer your question if you want your value to be an optional and possible empty value declare v as the following var v:String? if you would like it to be a non-optional value with a value always stored within v you should declare it as the following var v = "". Swift will interpret this declaration as a String.
To answer your second question, defining without values in swift is 100% allowed, it really just depends on how you want to handle your code. I use optional values in my code all the time, but I don't necessarily like optionals so i usually like to declare my values such as var v = "" that way if the value is empty my UI or whatever else i'm manipulating won't break. But if i need to ensure a value is present i will have to make my value optional so i can use an if statement to check whether it's a valid value or not.
Shorter version of what I'm trying to say is, you are receiving the compiler warning because you are declaring v as a non-optional value rather than an optional value.
I've been trying to get this function to return a Bool value but I don't understand why i'm getting the error "missing return in a function expected to return 'Bool'. I've been looking around online and tried different things but I can't seem to find a solution. Any help would be appreciated!
func trueSquare(a:[Int], b:[Int]) -> Bool {
for i in b[0]...b.endIndex {
if b[i] == a[i]*a[i] {
return true
}
else {
return false
}
}
}
EDIT: I have changed the loop to for i in 0...(b.count - 1) but I am still getting the same error even when I call the function with a and b both having the same numbers of elements.
What if your array has no element? Then for each loop never runs and then your method returns nothing which is obviously wrong. So you need to return value even outside of the loop.
But, your logic is bad. You're returning boolean value depending on if just first element from b is equal to a*a.
So, logic should be something like: if every element meets the condition, then return true, otherwise, return false. To achieve this, in Swift 4.2+ you can use method allSatisfy
func trueSquare(a:[Int], b:[Int]) -> Bool {
guard a.count == b.count else { return false } // if arrays have different number of elements, return false
return a.enumerated().allSatisfy {$0.element * $0.element == b[$0.offset]}
}
I suspect the compiler requires a return value for the case when the loop is not executed at all.
Now, a ClosedRange can never be empty, so b[0]...b.endIndex won't ever be empty (if it results in an empty or invalid range, the code would crash), but the compiler is not smart enough to know that.
PS: Are you sure b[0]...b.endIndex is actually the sequence you want to loop over. This creates a range from the first element of b to the endIndex of b. That doesn't make any sense to me.
Your function does not handle case where b is an empty array.
You need to define what you want the return value to be for such case, because your loop will be skipped when b is an empty array.
Secondly, your logic is also incomplete, because if the condition is good for i==0, you immediately return true, without checking the rest of the items.
Thirdly, you probably want to make sure a and b have same length.
So here is what your function should look like:
func trueSquare(a:[Int], b:[Int]) -> Bool {
if a.count != b.count {
return false
}
for i in 0..<b.count {
if b[i] != a[i]*a[i] {
return false
}
}
return true
}
I'm building a didactic compiler, and I'd like to check if the function will always return a value. I intend to do this in the semantic analysis step (as this is not covered by the language grammar).
Out of all the flow control statements, this didactic language only has if, else, and while statements (so no do while, for, switch cases, etc). Note that else if is also possible. The following are all valid example snippets:
a)
if (condition) {
// non-returning commands
}
return value
b)
if (condition) {
return value
}
return anotherValue
c)
if (condition) {
return value1
} else {
return value2
}
// No return value needed here
I've searched a lot about this but couldn't find a pseudoalgorithm that I could comprehend. I've searched for software path testing, the white box testing, and also other related Stack Overflow questions like this and this.
I've heard that this can be solved using graphs, and also using a stack, but I have no idea how to implement those strategies.
Any help with pseudocode would be very helpful!
(and if it matters, I'm implementing my compiler in Swift)
If you have a control flow graph, checking that a function always returns is as easy as checking that the implicit return at the end of the function is unreachable. So since there are plenty of analyses and optimizations where you'll want a CFG, it would not be a bad idea to construct one.
That said, even without a control flow graph, this property is pretty straight forward to check assuming some common restrictions (specifically that you're okay with something like if(cond) return x; if(!cond) return y; being seen as falling of the end even though it's equivalent to if(cond) return x; else return y;, which would be allowed). I also assume there's no goto because you didn't list it in your list of control flow statements (I make no assumptions about break and continue because those only appear within loops and loops don't matter).
We just need to consider the cases of what a legal block (i.e. one that always reaches a return) would look like:
So an empty block would clearly not be allowed because it can't reach a return if it's empty. A block that directly (i.e. not inside an if or loop) contains a return would be allowed (and if it isn't at the end of the block, everything after the return in the block would be unreachable, which you might also want to turn into an error or warning).
Loops don't matter. That is, if your block contains a loop, it still has to have a return outside of the loop even if the loop contains a return because the loop condition may be false, so there's no need for us to even check what's inside the loop. This wouldn't be true for do-while loops, but you don't have those.
If the block directly contains an if with an else and both the then-block and the else-block always reach a return, this block also always reaches a return. In that case, everything after the if-else is unreachable. Otherwise the if doesn't matter just like loops.
So in pseudo code that would be:
alwaysReturns( {} ) = false
alwaysReturns( {return exp; ...rest} ) = true
alwaysReturns( { if(exp) thenBlock else elseBlock; ...rest}) =
(alwaysReturns(thenBlock) && alwaysReturns(elseBlock)) || alwaysReturns(rest)
alwaysReturns( {otherStatement; ...rest} ) = alwaysReturns(rest)
So, after 5 hours thinking how to implement this, I came up with a decent solution (at least I haven't been able to break it so far). I actually spent most of the time browsing the web (with no luck) than actually thinking about the problem and trying to solve it on my own.
Below is my implementation (in Swift 4.2, but the syntax is fairly easy to pick up), using a graph:
final class SemanticAnalyzer {
private var currentNode: Node!
private var rootNode: Node!
final class Node {
var nodes: [Node] = []
var returnsExplicitly = false
let parent: Node?
var elseNode: Node!
var alwaysReturns: Bool { return returnsExplicitly || elseNode?.validate() == true }
init(parent: Node?) {
self.parent = parent
}
func validate() -> Bool {
if alwaysReturns {
return true
} else {
return nodes.isEmpty ? false : nodes.allSatisfy { $0.alwaysReturns }
}
}
}
/// Initializes the components of the semantic analyzer.
func startAnalyzing() {
rootNode = Node(parent: nil)
currentNode = rootNode
}
/// Execute when an `if` statement is found.
func handleIfStatementFound() {
let ifNode = Node(parent: currentNode)
let elseNode = Node(parent: currentNode)
// Assigning is not necessary if the current node returns explicitly.
// But assigning is not allowed if the else node always returns, so we check if the current node always returns.
if !currentNode.alwaysReturns {
currentNode.elseNode = elseNode
}
currentNode.nodes += [ ifNode, elseNode ]
currentNode = ifNode
}
/// Execute when an `else` statement is found.
func handleElseStatementFound() {
currentNode = currentNode.elseNode
}
/// Execute when a branch scope is closed.
func handleBranchClosing() {
currentNode = currentNode.parent! // If we're in a branch, the parent node is never nil
}
/// Execute when a function return statement is found.
func handleReturnStatementFound() {
currentNode.returnsExplicitly = true
}
/// Determine whether the function analyzed always returns a value.
///
/// - Returns: whether the root node validates.
func validate() -> Bool {
return rootNode.validate()
}
}
Basically what it does is:
When it finds an if statement is create 2 new nodes and point the current node to both of them (as in a binary tree node).
When the else statement is found, we just switch the current node to the else node created previously in the if statement.
When a branch is closed (e.g. in an if statement's } character), it switches the current node to the parent node.
When it finds a function return statement, it can assume that the current node will always have a return value.
Finally, to validate a node, either the node has an explicit return value, or all of the nodes must be valid.
This works with nested if/else statements, as well as branches without return values at all.
Is there a way in Swift to assign conditional expressions similar to this
let foo = if (bar == 2) {
100
} else {
120
}
(or with a switch case).
(Don't want to have to use ternary operator for this).
This kind of assignement is good for functional style / immutability. The expressions have a return value in this case.
Note: it's a general question, this is just simplified example, imagine e.g. a switch case with a lot of values, pattern matching, etc. You can't do that with ternary operator.
Btw also note that there are languages that don't support ternary operator because if else returns a value, so it's not necessary, see e.g. Scala.
You can use a closure to initialize an immutable:
let foo: Int = {
if bar == 2 {
return 100
} else {
return 120
}
}()
The advantage of using a closure is that it's a function, so you can use any complex logic inside, implemented in a clean way and not through nested ternary operators. It can be a switch statement, it can be obtained as the return value of a function followed by some calculations, it can be a pattern matching case, it can be a combination of them all, etc.
Said in other words, it's the same as initializing with the return value of a function, but the difference is that the function is inline and not somewhere else, with readability benefits.
Just for completeness, if the variable is mutable, you can use deferred initialization:
var foo: Int
// Any statement here
if bar == 2 {
foo = 100
} else {
foo = 120
}
// Other statements here
myFunc(foo)
so you can declare a mutable variable, and initialize it anywhere in the same scope, but before using it the variable must be initialized.
Update: Since Swift 2.0, deferred initialization also works with immutables.
Let's say I have function which returns optional. nil if error and value if success:
func foo() -> Bar? { ... }
I can use following code to work with this function:
let fooResultOpt = foo()
if let fooResult = fooResultOpt {
// continue correct operations here
} else {
// handle error
}
However there are few problems with this approach for any non-trivial code:
Error handling performed in the end and it's easy to miss something. It's much better, when error handling code follows function call.
Correct operations code is indented by one level. If we have another function to call, we have to indent one more time.
With C one usually could write something like this:
Bar *fooResult = foo();
if (fooResult == null) {
// handle error and return
}
// continue correct operations here
I found two ways to achieve similar code style with Swift, but I don't like either.
let fooResultOpt = foo()
if fooResult == nil {
// handle error and return
}
// use fooResultOpt! from here
let fooResult = fooResultOpt! // or define another variable
If I'll write "!" everywhere, it just looks bad for my taste. I could introduce another variable, but that doesn't look good either. Ideally I would like to see the following:
if !let fooResult = foo() {
// handle error and return
}
// fooResult has Bar type and can be used in the top level
Did I miss something in the specification or is there some another way to write good looking Swift code?
Your assumptions are correct—there isn't a "negated if-let" syntax in Swift.
I suspect one reason for that might be grammar integrity. Throughout Swift (and commonly in other C-inspired languages), if you have a statement that can bind local symbols (i.e. name new variables and give them values) and that can have a block body (e.g. if, while, for), those bindings are scoped to said block. Letting a block statement bind symbols to its enclosing scope instead would be inconsistent.
It's still a reasonable thing to think about, though — I'd recommend filing a bug and seeing what Apple does about it.
This is what pattern matching is all about, and is the tool meant for this job:
let x: String? = "Yes"
switch x {
case .Some(let value):
println("I have a value: \(value)")
case .None:
println("I'm empty")
}
The if-let form is just a convenience for when you don't need both legs.
If what you are writing is a set of functions performing the same sequence of transformation, such as when processing a result returned by a REST call (check for response not nil, check status, check for app/server error, parse response, etc.), what I would do is create a pipeline that at each steps transforms the input data, and at the end returns either nil or a transformed result of a certain type.
I chose the >>> custom operator, that visually indicates the data flow, but of course feel free to choose your own:
infix operator >>> { associativity left }
func >>> <T, V> (params: T?, next: T -> V?) -> V? {
if let params = params {
return next(params)
}
return nil
}
The operator is a function that receives as input a value of a certain type, and a closure that transforms the value into a value of another type. If the value is not nil, the function invokes the closure, passing the value, and returns its return value. If the value is nil, then the operator returns nil.
An example is probably needed, so let's suppose I have an array of integers, and I want to perform the following operations in sequence:
sum all elements of the array
calculate the power of 2
divide by 5 and return the integer part and the remainder
sum the above 2 numbers together
These are the 4 functions:
func sumArray(array: [Int]?) -> Int? {
if let array = array {
return array.reduce(0, combine: +)
}
return nil
}
func powerOf2(num: Int?) -> Int? {
if let num = num {
return num * num
}
return nil
}
func module5(num: Int?) -> (Int, Int)? {
if let num = num {
return (num / 5, num % 5)
}
return nil
}
func sum(params: (num1: Int, num2: Int)?) -> Int? {
if let params = params {
return params.num1 + params.num2
}
return nil
}
and this is how I would use:
let res: Int? = [1, 2, 3] >>> sumArray >>> powerOf2 >>> module5 >>> sum
The result of this expression is either nil or a value of the type as defined in the last function of the pipeline, which in the above example is an Int.
If you need to do better error handling, you can define an enum like this:
enum Result<T> {
case Value(T)
case Error(MyErrorType)
}
and replace all optionals in the above functions with Result<T>, returning Result.Error() instead of nil.
I've found a way that looks better than alternatives, but it uses language features in unrecommended way.
Example using code from the question:
let fooResult: Bar! = foo();
if fooResult == nil {
// handle error and return
}
// continue correct operations here
fooResult might be used as normal variable and it's not needed to use "?" or "!" suffixes.
Apple documentation says:
Implicitly unwrapped optionals are useful when an optional’s value is confirmed to exist immediately after the optional is first defined and can definitely be assumed to exist at every point thereafter. The primary use of implicitly unwrapped optionals in Swift is during class initialization, as described in Unowned References and Implicitly Unwrapped Optional Properties.
How about the following:
func foo(i:Int) ->Int? {
switch i {
case 0: return 0
case 1: return 1
default: return nil
}
}
var error:Int {
println("Error")
return 99
}
for i in 0...2 {
var bob:Int = foo(i) ?? error
println("\(i) produces \(bob)")
}
Results in the following output:
0 produces 0
1 produces 1
Error
2 produces 99