In an e file, it's perfectly legal to say:
print 5;
But at the same time, the following thing doesn't work:
type some_type_e : [ VAL1, VAL2 ];
print VAL2; // issues a compile error
The parser expects VAL2 to be a variable name and doesn't interpret it as a constant.
At the same time, this is allowed:
var some_int : int = 10;
if some_int != 5 {
print "Some int not 5"
};
var some_enum : some_type_e = VAL1;
if some_enum != VAL2 {
print "Some enum not VAL2";
};
In this case, VAL2 is interpreted as a constant.
The (simplified) syntax for print is print <exp>. Is an enum literal not an expression?
I've figured it out. VAL2 on its own is meaningless as a constant. It has to be qualified with the type name. The proper way to do it is:
print some_type_e'VAL2;
Related
I'm trying to make a type that should represent a "slice" of some indexable collection.
I know that there are some similar types in F# but not one that specifies the criteria that I need.
To do this it needs to carry a reference to the collection of type 'content and the content needs to be indexable. So I tried this constraint since a type only needs to have the member Item (get/set) so I tried this
type Slice<'a, 'content when 'content: (member Item: int -> 'a)>
This still throw the usual error
So is it possible to constrain a type to still be generic but constraint to be indexable?
I think something like this should work:
type Slice<'a, 'content when 'content: (member get_Item: int -> 'a)> =
{
Content : 'content
Start : int
Stop : int
}
with
member inline slice.get_Item(i) =
slice.Content.get_Item(slice.Start + i)
I've implemented get_Item on Slice as well, so you can take a slice of a slice. Here are some values of this type:
let strSlice =
{
Content = "hello"
Start = 1
Stop = 2
}
let arraySlice =
{
Content = [| 2; 4; 6; 8 |]
Start = 0
Stop = 3
}
let strSliceSlice =
{
Content = strSlice
Start = 0
Stop = 1
}
[<Interface>]
type Indexable<'a> =
abstract member Item: int -> 'a with get
[<Struct>]
type Slice<'a> =
{
content: Indexable<'a>
start: int
len: int
}
with
interface Indexable<'a> with
member I.Item with get(i) = I.[idx]
member S.Item with get(idx) =
if idx >= S.len
then raise(IndexOutOfRangeException())
else S.content.[S.start+idx]
This works.
I want to compare fields of a JSONObject.
Here is the code I am trying:
x = JSONObject()
x.put("num",1)
if(x["num"] as Int < 2){
print("Good work!")
}
Error shown by IDE:
Type expected
Expecting a '>
You need to put x["num"] as Int into parenthesis, otherwise the compiler will think that you are trying to do Int < 2 which doesn't really make any sense.
val x = JSONObject()
x.put("num", 1)
if ((x["num"] as Int) < 2) {
print("Good work!")
}
But btw, this will throw a ClassCastException if the value is not actually an Int. It is in your minimal case, but you don't always actually know. In that scenario I would do something like this:
val x = JSONObject()
x.put("num", "b")
val num = x["num"]
if (num is Int && num < 2) {
print("Good work!")
}
The current code doesn't work as-is because the compiler sees Int<42 and thinks that you're trying to use a type parameter for the Int type (like Int<T>) which obviously doesn't work for 2 reasons:
Type expected because 42 is not a type
Expecting a '>' because well, there is no >
As #Alex.T said, you can disambiguate this by putting (..) around the cast expression: (x["num"] as Int) < 2.
However, if you're OK with your expression failing at runtime like this, you should instead use a typed getter like getInt():
val x = JSONObject()
x.put("num", 1)
if (x.getInt("num") < 2) {
print("Good work!")
}
If the value is not an integer, this will fail at runtime with org.json.JSONException (as mentioned in the Javadoc linked above). You can catch this exception if you want to handle the error gracefully.
Alternatively you can use a safer option with a default value, like optInt(), which will default to the given value if the key is not present in the JSON OR if the value is not a number.
I am trying to find the find the result of num1 raised to the power num2:
This is my code ->
fun power(num1 : Int, num2: Int): Int {
var result = 1
while (num2 != 0) {
return result *= num1
num2--
}
}
But the above code is producing the following error -->
Calculator.kt:30:16: error: assignments are not expressions, and only expressions are allowed in this context
return result *= num1
^
Calculator.kt:33:5: error: a 'return' expression required in a function with a block body ('{...}')
}
^
I have read a number of articles but not able to understand. Any help will be highly appreciated.
Thank you
An expression is something that evaluates to a value. An assignment is something that assigns a value to a variable or property.
x *= y is an assignment that is shorthand for x = x * y.
You cannot return an assignment, because it does not evaluate to a value. An assignment contains an expression on the right side of the equals sign, but as a whole does not represent an expression.
There are some other syntax problems you have. You can't modify a function paramter's value (num2-- isn't allowed).
The logic also doesn't make sense. return returns an expression immediately. To fix your code, you need to create a local variable from num2, and move the return statement to the end.
fun power(num1 : Int, num2: Int): Int {
var result = 1
var count = num2
while (count != 0) {
result *= num1
count--
}
return result
}
FYI, there's a function called repeat that is simpler than using a while loop with a counter. It runs the code inside the brackets by the number of times you give.
fun power(num1 : Int, num2: Int): Int {
var result = 1
repeat(num2) {
result *= num1
}
return result
}
You function contains multiple errors, I suggest you to study Kotlin, here a reference. Kotlin website has some more material.
Back to your problem, I have modified your function:
fun power(num1 : Int, num2: Int): Int {
var result = 1
var num2_local = num2
while (num2_local != 0) {
result *= num1
num2_local--
}
return result
}
Problems with your version:
you return from the function immediately
basic types args passed to kotlin functions are passed by const copy, this means that you cannot modify them (e.g num2--)
If you keep a local modifiable reference (var keyword) withing your function, then you achieve your goal
Here's the problem in which I encountered this issue:
The function should compare the value at each index position and score a point if the value for that position is higher. No point if they are the same. Given a = [1, 1, 1] b = [1, 0, 0] output should be [2, 0]
fun compareArrays(a: Array<Int>, b: Array<Int>): Array<Int> {
var aRetVal:Int = 0
var bRetVal:Int = 0
for(i in 0..2){
when {
a[i] > b[i] -> aRetVal + 1 // This does not add 1 to the variable
b[i] > a[i] -> bRetVal++ // This does...
}
}
return arrayOf(aRetVal, bRetVal)
}
The IDE even says that aRetVal is unmodified and should be declared as a val
What others said is true, but in Kotlin there's more. ++ is just syntactic sugar and under the hood it will call inc() on that variable. The same applies to --, which causes dec() to be invoked (see documentation). In other words a++ is equivalent to a.inc() (for Int or other primitive types that gets optimised by the compiler and increment happens without any method call) followed by a reassignment of a to the incremented value.
As a bonus, consider the following code:
fun main() {
var i = 0
val x = when {
i < 5 -> i++
else -> -1
}
println(x) // prints 0
println(i) // prints 1
val y = when {
i < 5 -> ++i
else -> -1
}
println(y) // prints 2
println(i) // prints 2
}
The explanation for that comes from the documentation I linked above:
The compiler performs the following steps for resolution of an operator in the postfix form, e.g. a++:
Store the initial value of a to a temporary storage a0;
Assign the result of a.inc() to a;
Return a0 as a result of the expression.
...
For the prefix forms ++a and --a resolution works the same way, and the effect is:
Assign the result of a.inc() to a;
Return the new value of a as a result of the expression.
Because
variable++ is shortcut for variable = variable + 1 (i.e. with assignment)
and
variable + 1 is "shortcut" for variable + 1 (i.e. without assignment, and actually not a shortcut at all).
That is because what notation a++ does is actually a=a+1, not just a+1. As you can see, a+1 will return a value that is bigger by one than a, but not overwrite a itself.
Hope this helps. Cheers!
The equivalent to a++ is a = a + 1, you have to do a reassignment which the inc operator does as well.
This is not related to Kotlin but a thing you'll find in pretty much any other language
I am making a program that solves a math expression, for example, 2+2. Can I set an integer equal to something like this:
val input = "2+2"
input.toInt()
Kotlin doesn't have any built in ways for evaluating arbitrary expressions. The toInt function can only parse a String containing a single whole number (it's just a wrapper for Integer.parseInt).
If you need this functionality, you'll have to parse and evaluate the expression yourself. This problem is no different than having to do it in Java, for which you can find discussion and multiple solutions (including hacks, code samples, and libraries) here.
No you cannot convert directly a String Mathematical Expression to Integer.
But you can try following approach to convert String Mathematical Expression to Integer ->>
var exp: String = "2+3-1*6/4"
var num: String = ""
var symbol: Char = '+'
var result: Int = 0
for(i in exp)
{
if(i in '0'..'9')
num += i
else
{
if(symbol == '+')
result += Integer.parseInt(num)
else if(symbol == '-')
result -= Integer.parseInt(num)
else if(symbol == '*')
result *= Integer.parseInt(num)
else if(symbol == '/')
result /= Integer.parseInt(num)
num=""
symbol = i
}
}
//To calculate the divide by 4 ( result/4 ) in this case
if(symbol == '+')
result += Integer.parseInt(num)
else if(symbol == '-')
result -= Integer.parseInt(num)
else if(symbol == '*')
result *= Integer.parseInt(num)
else if(symbol == '/')
result /= Integer.parseInt(num)
println("result is $result") //Output=> result is 6
}
No you can't.
You can like this:
val a = "2"
val b = "2"
val c = a.toInt() + b.toInt()
Or
val input = "2+2"
val s = input.split("+")
val result = s[0].toInt() + s[1].toInt()
This can be done with the kotlin script engine. For details see Dynamically evaluating templated Strings in Kotlin
But in a nutshell it's like this:
val engine = ScriptEngineManager().getEngineByExtension("kts")!!
engine.eval("val x = 3")
val res = engine.eval("x + 2")
Assert.assertEquals(5, res)
No, Integer cannot be equal to math expression.
You may use String Templates
Strings may contain template expressions, i.e. pieces of code that are evaluated and whose results are concatenated into the
string.
A template expression starts with a dollar sign ($) and consists of either a simple name:
val i = 10
val s = "i = $i" // evaluates to "i = 10"