The following example of Kotlin source code returns an error when compiled:
fun main() {
var index: Int // create an integer used to call an index of an array
val myArray = Array(5) {i -> i + 1} // create an array to call from
val condition = true // makes an if statement run true later
if (condition) {
index = 2 // sets index to 2
}
println( myArray[index] ) // should print 2; errors
}
The error says that the example did not initialize the variable index by the time it is called, even though it is guaranteed to initialize within the if statement. I understand that this problem is easily solved by initializing index to anything before the if statement, but why does the compiler not initialize it? I also understand that Kotlin is still in beta; is this a bug, or is it intentional? Finally, I am using Replit as an online IDE; is there a chance that the compiler on the website simply is an outdated compiler?
The compiler checks whether there is a path in your code that the index may not be initialized based on all the path available in your code apart from the value of the parameters. You have an if statement without any else. If you add the else statement you will not get any compile error.
Related
This question already has answers here:
What is the return type of the indexing operation?
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Closed 2 years ago.
I'm currently learning Rust coming from JavaScript.
My problem is the following:
fn main() {
let name = String::from("Tom");
let sliced = name[..2];
println!("{}, {}", name, sliced);
}
This doesn't work. Saying "doesn't have a size known at compile-time".
To fix this I need to add & the referencing operator.
fn main() {
let name = String::from("Tom");
let sliced = &name[..2];
println!("{}, {}", name, sliced);
}
I know I need to add & before name and & is the referencing operator. But I just don't know why I actually need to do that?
By referencing a variable the reference refers to variable name but does not own it. The original value will not get dropped if my reference gets out of scope. Does that mean that the variable gets out of scope if i do name[...] and the variable gets dropped and because of that i need to create a reference to it to prevent that?
Could somebody explain me that?
I know I need to add & before name and & is the referencing operator. But I just don't know why I actually need to do that.
I understand where the confusion come from, because when you look at index() it returns &Self::Output. So it already returns a reference, what's going on?
It's because the indexing operator is syntactic sugar and uses the Index trait. However, while it uses index() which does return a reference, that is not how it is desugared.
In short x[i] is not translated into x.index(i), but actually to *x.index(i), so the reference is immediately dereferenced. That's how you end up with a str instead of a &str.
let foo = "foo bar"[..3]; // str
// same as
let foo = *"foo bar".index(..3); // str
That's why you need to add the & to get it "back" to a reference.
let foo = &"foo bar"[..3]; // &str
// same as
let foo = &*"foo bar".index(..3); // &str
Alternatively, if you call index() directly, then it isn't implicitly dereferenced of course.
use std::ops::Index;
let foo = "foo bar".index(..3); // &str
Trait std::ops::Index - Rust Documentation:
container[index] is actually syntactic sugar for *container.index(index)
The same applies to IndexMut.
class Solution {
val message: String //error : val must be initialized or abstract
message = "love" //error : val cannot be reassigned
}
I understand what's happening in here - val cannot be reassigned.
So when I need val but can not initialize it i used to use by lazy
class Solution {
fun love(){
val message : String
message = "love" //this works
message = "hate" //this is error "val cannot be reassigned"
}
}
Here I can delcare val without initialization and later write codemessage = "love".what's happening here?
#deHaar noticed correctly that only var (mutable variable) is appropriate in your case.
The error you get is absolutely correct and expected.
what's happening here?
When you declare a read-only variable without initializing it you have to make sure that each execution path will have a value in this read-only variable. It means that Kotlin makes sure if your read-only variable was or was not initialized in every place it is used and raises errors if the variable is used inappropriately.
Here you have only one execution path as there are no when or if statements that can split execution into several possible paths.
class Solution {
fun love(){
val message : String
message = "love" // Kotlin knows that `message` was not yet initialized
message = "hate" // Kotlin knows that `message` was yet initialized! It does not allow to modify the value.
}
}
Here is what Kotlin documentation says:
... it is also possible (but discouraged) to split the declaration and the initial assignment, and even to initialize in multiple places based on some condition. You can only read the variable at a point where the compiler can prove that every possible execution path will have initialized it. If you're creating a read-only variable in this way, you must also ensure that every possible execution path assigns to it exactly once.
Example of an execution path
Using when or if statement you create two or more execution paths. Execution paths can be presented as a graph, I'll use #number as a node number. Example:
class Solution {
fun love(){
// #1
val message : String
if (System.currentTimeMillisec() % 2 == 0) {
message = "Not empty"
// #2
}
if (message.isEmpty) { // Error! Message could be not initialized at this point!
println("Empty message")
// #3
}
}
}
Looking at this example, that does not compile, we can calculate at least 3 execution paths.
#1 (none of the if statements was entered. All conditions are false)
#1 -> #2
#1 -> #3
Kotlin can calculate these paths and check if the message variable is initialized in every path it is used. As we can see, as soon as you reach the evaluation of the second if statement (in case of first and third paths) your program will crash because the message has no value. It has no address in memory and a computer which runs this program does not know how to get a value from an address that does not exist.
Now, let's modify this code to make it work:
class Solution {
fun love(){
// #1
val message : String
if (System.currentTimeMillisec() % 2 == 0) {
message = "Not empty"
// #2
} else {
message = ""
// #3
}
if (message.isEmpty) { // Error! Message could be not initialized at this point!
println("Empty message")
// #4
}
}
}
Execution paths:
#1 -> #2
#1 -> #3 -> #4
In this example, Kotlin is sure that the message read-only variable is initialized because there is a 100% chance that one of node 2 or node 3 will be executed. Right after the line where the message gets its initial value (initialized) Kotlin treats this variable as a read-only variable with a value.
Questions are welcome. I will try to simplify this answer.
Is there an operator that allows to process result/success whether or not Mono is empty. For example:
Mono<Bar> result = sourceMono.flatMap(n -> process(n)).switchIfEmpty(process(null));
where:
Mono<Bar> process(Foo in){
Optional<Foo> foo = Optional.ofNullable(in);
...
}
is there a shortcut operator that allows something like below or similar?
Mono<Bar> result = sourceMono.shortCut(process);
More specifically, mono.someOperator() returns Optional<Foo> which would contain null when Mono is empty and have value otherwise.
I wanted to avoid to create process method as mentioned above and just have a block of code but not sure which operator can help without duplicating block.
There is no built-in operator to do exactly what you want.
As a workaround, you can convert the Mono<Foo> to a Mono<Optional<Foo>> that emits an empty Optional<Foo> rather than completing empty, and then operate on the emitted Optional<Foo>.
For example:
Mono<Bar> result = fooMono // Mono<Foo>
.map(Optional::of) // Mono<Optional<Foo>> that can complete empty
.defaultIfEmpty(Optional.empty()) // Mono<Optional<Foo>> that emits an empty Optional<Foo> rather than completing empty
.flatMap(optionalFoo -> process(optionalFoo.orElse(null)));
As per above #phil's workaround, here is a reusable function:
private final <T> Mono<Optional<T>> afterSucess(Mono<T> source) {
return source
.map(Optional::of) //
.defaultIfEmpty(Optional.empty());
}
then invoke in publisher line:
Foo<Bar> result = fooMono
.transformDeferred(this::afterSucess)
.flatMap(optionalFoo -> process(optionalFoo.orElse(null)));
I have the following function:
override fun countForTicket(dbc: SQLiteDatabase, ticketId: Long): Int {
var ret: Int
dbc.query(
TABLE_SECOND_CHANCE_PRIZES, arrayOf("count(id)"),
"ticket = ?", arrayOf(ticketId.toString()),
null, null, null
).use { c ->
ret = if (c.moveToFirst()) {
c.getInt(0)
} else {
0
}
}
return ret
}
The problem is that in line return ret ret is underlined with red and when trying to compile it gives me error:
Variable 'ret' must be initialized
From my point of view it seems that ret is always initialized. What am I missing?
Is it because the initialization is happening in a lambda and the compiler cannot guarantee that the variable is initialized?
The compiler isn't smart enough to know for sure the lambda will be run once, so it can't figure this out for you.
The reason we don't have this problem with many of the standard library higher-order functions is that they utilize contracts, which tell the compiler what they are doing with the lambda that is passed in (such as guaranteeing that the lambda will be called exactly once).
Unfortunately, Closeable.use() doesn't specify a contract (possibly because of it re-throwing exceptions?).
But use does return the result of calling the lambda, so you could do
val ret = dbc.query(...).use { c ->
if (c.moveToFirst()) {
c.getInt(0)
} else {
0
}
}
The compiler don't allow unsafe variables like that to be returned. A variable must always be something.
In your case, ret is initialized inside a lambda. The compiler doesn't know if this lambda is executed or not. If not, ret remains in its unsafe state. Throwing a NullPointerException at the end.
If you're sure that this variable is always assigned you can look at lateinit variables. You can also put a default value to it var ret = 0 and ommit the else statement.
The following code works correctly - output: You chose Test 1
package main
import (
"fmt"
)
type TNameMap map[int]string
var nameMap TNameMap
func init() {
nameMap = make(TNameMap)
nameMap[1] = "You chose Test 1"
nameMap[2] = "You chose Test 2"
nameMap[3] = "You chose Test 3"
}
func main() {
fmt.Println(nameMap[1])
}
If I comment out the first line in init() i.e //nameMap = make(TNameMap) , I get a panic when main() runs, because nameMap was never initialized:
panic: runtime error: assignment to entry in nil map
But - if in init() I write nameMap := make(TNameMap)
instead of nameMap = make(TNameMap) , I get no panic, but also no output - main() simply runs and process terminates.
I understand that if I use the Initialization operator - nameMap := make(TNameMap) - I have declared a new variable nameMap that is scoped only to the init() function and so only the package level variable var nameMap TNameMap is in scope for main(), resulting in no output, because the package level var holds no map data.
But, I am confused: Why don't I get the panic in that situation? If main() is making the call on the package var, it was never initialized - so why no panic?
According to the Go spec:
A nil map is equivalent to an empty map except that no elements may be
added.
This means that you can read from a nil map, but not write. Just like the panic says "assignment to entry in nil map". If you comment out just the line nameMap = make(TNameMap) it will crash because you attempt to write to it in init (which is where the panic happens). If you comment out the entirety of init the Println will not crash because you're permitted to access (read from) a nil map.
Changing the assignment to a declaration is just masking the real issue here, what's happening is it's making all the assignments valid, and then discarding the result. As long as you make the assignments valid (either by removing them or making a temporary variable), then you will observe the same behavior in Println.
The value returned by a nil map is always the zero value of the value type of the map. So a map[T]string returns "", a map[T]int returns 0, and so on. (Of course, if you check with val,ok := nilMap[key] then ok will be false).