I currently start to learn Kotlin and I was making this code
val a = "1"
val b = a[0]
val c = b.toInt()
println(c)
When I run the code, the result is 49. What really happened? Because I think the result will be 1.
a is a String, which is a CharSequence. That is why you can access a[0] in the first place. a.get(0) or a[0] then returns a Char. Char on the other hand returns its character value when calling toInt(), check also the documentation of toInt().
So your code commented:
val a = "1" // a is a String
val b = a[0] // b is a Char
val c = b.toInt() // c is (an Int representing) the character value of b
If you just want to return the number you rather need to parse it or use any of the answers you like the most of: How do I convert a Char to Int?
(one simple way being b.toString().toInt()).
a is String,
when you get a[index] return type is char,
in kotlin char.toInt method return ASCII code of the character and it's 49
if you want to get the integer value of "1" just use toString method
val a = "1"
val b = a[0].toString()
val c = b.toInt()
println(c)
prints:1
In your example a is a String, but String. String is under the hood an Array of Char. And by accessing your String using a[0] operator, you get first element of this Char Array. So you get Char '1', not String "1". And now, when you run '1'.toInt() function on Char - it will return ASCII code of that Char. When you run "1".toInt() on String - it will convert this String into Int "1". When you need to get Int value of first letter in your String, you need to convert it first into String:
val a = "123"
val b = a[0].toString() // returns first Char of String "123" and converts to String
val c = b.toInt() // returns Int: 1
or in one line:
"123"[0].toString().toInt()
Related
Here is the code. n continually outputs 50 and not 2:
import java.util.*
fun main(args: Array<String>) {
val scanner = Scanner(System.`in`)
val n = scanner.next().first().toInt()
val array1 = readLine()!!.split(" ").map { it.toInt() }
var product:Int=0
println(n)
println(array1[0])
println(array1[1])
if (n ==2) {
product = array1[0] * array1[1]
}
println(product)
}
Sample Input:
2
5 3
Output:
2
5 3
50
5
3
0
How do I use scanner in kotlin to read in 2 lines?
Short form
Use scanner.nextInt() instead of scanner.next().first().toInt().
Explanation
By calling scanner.next() you receive the next complete token of the Scanner as a String. Then you take the first character using first(). The problem is, that calling toInt() on a Char will not parse the string as an integer value but return the ASCII char code of the char.
Example: '2'.toInt() returns 50 and not 2 since the ASCII char code of 2 is 50. (See https://www.cs.cmu.edu/~pattis/15-1XX/common/handouts/ascii.html)
Conclusion: In your case, directly read the integer value from the Scanner using scanner.nextInt(), but if you want to convert a char to an integer by "parsing", convert it to a string first: '2'.toString().toInt() will return an integer with the value 2
Addition
It is probably nice to know that you can also use this the other way round: 50.toChar() returns a character with the value '2'.
👍
I'm trying to find the simplest way to convert a digit (0..9) into the respective character '0'..'9' in Kotlin.
My initial attempt was to write the following code:
fun convertToCharacter() {
val number = 0
val character = number.toChar()
println(character)
}
Of course, after running, I quickly saw that this produces \u0000, and not '0' like I expected. Then, remembering from how to do this in Java, I modified the code to add '0', but then this would not compile.
fun convertToCharacter() {
val number = 0
val character = number.toChar() + '0'
println(character)
}
What is the appropriate way to convert a number into its respective character counterpart in Kotlin? Ideally, I'm trying to avoid pulling up the ASCII table to accomplish this (I know I can add 48 to the number since 48 -> '0' in ASCII).
val character = '0' + number
is the shortest way, given that the number is in range 0..9
Kotlin stdlib provides this function since 1.5.0.
fun Int.digitToChar(): Char
Returns the Char that represents this decimal digit. Throws an exception if this value is not in the range 0..9.
If this value is in 0..9, the decimal digit Char with code '0'.code + this is returned.
Example
println(5.digitToChar()) // 5
println(3.digitToChar(radix = 8)) // 3
println(10.digitToChar(radix = 16)) // A
println(20.digitToChar(radix = 36)) // K
Like you said, probably the easiest way to convert an Int to the Char representation of that same digit is to add 48 and call toChar():
val number = 3
val character = (number + 48).toChar()
println(character) // prints 3
If you don't want to have the magic 48 number in your program, you could first parse the number to a String and then use toCharArray()[0] to get the Char representation:
val number = 3
val character = number.toString().toCharArray()[0]
println(character) // prints 3
Edit: in the spirit of the attempt in your question, you can do math with '0'.toInt() and get the result you were expecting:
val number = 7
val character = (number + '0'.toInt()).toChar()
println(number) // prints 7
How about 0.toString() instead of 0.toChar() ? If you are specifically after single digits, then 0.toString()[0] will give you a Char type
You can use an extension like this:
fun Int.toReadableChar(): Char {
return ('0'.toInt() + this).toChar()
}
You can apply this to any other class you want :)
Example:
println(7.toReadableChar())
>> 7
How to convert a string value which contains float representation to integer in kotlin?
I tried to convert string to float with .toFloat() and then converted it to an integer using toInt() and it works flawlessly.
But how to convert such string to integer directly?
val strDemo = "42.22"
val intDemo = strDemo.toInt()
snippet above throws NumberFormatException because it is not correct number representaion of Integer.
But, when I try
val strDemo = "42.22"
val intDemo = strDemo.toFloat().toInt()
it converts the data with no exception because string gets converted to float first. And there is a correct number representation for a Float value.
Now how to bypass the toFloat() method and convert strDemo to Integer directly?
There's no magic function that will convert a decimal/float string numbers to integer directly. It has to be done this way. Even if you found one, I'm sure that the process toFloat().toInt() still happen on that function.
So the solution that you can do is to create an extension of String like this:
StringExt.kt
fun String.floatToInt(): Int {
return this.toFloat().toInt()
}
You can use it like this:
val strDemo = "42.22"
val intDemo = strDemo.floatToInt()
I am trying to get the MD5 format of string
Code:
fun getEncodedData(data: String): String? {
val MD5 = "MD5"
// Create MD5 Hash
val digest = java.security.MessageDigest
.getInstance(MD5)
digest.update(data.toByte())
val messageDigest = digest.digest()
// Create Hex String
val hexString = StringBuilder()
for (aMessageDigest in messageDigest) {
var h = Integer.toHexString(0xFF and aMessageDigest.toInt())
while (h.length < 2)
h = "0$h"
hexString.append(h)
}
return hexString.toString()
}
There is a crash at: digest.update(data.toByte()). I get number format Exception
Input I am passing for data: oEXm43
There is no crash if I pass ex: 11 as a string for input data
Should the input always should be integer in the string or can it be a mixture of number and characters.
You're trying to call the update method that takes a single byte parameter, and using toByte which converts the entire string's numerical value to a single byte. This conversion method is what fails on non-numerical values inside a String.
Instead, you can use the variant of update with a byte[] parameter, and convert your String to an array of bytes (one per character) with toByteArray:
digest.update(data.toByteArray())
I wanted to sum the digits of Long variable and add it to the variable it self, I came with the next working code:
private fun Long.sumDigits(): Long {
var n = this
this.toString().forEach { n += it.toString().toLong() }
return n
}
Usage: assert(48.toLong() == 42.toLong().sumDigits())
I had to use it.toString() in order to get it work, so I came with the next test and I don't get it's results:
#Test
fun toLongEquality() {
println("'4' as Long = " + '4'.toLong())
println("\"4\" as Long = " + "4".toLong())
println("\"42\" as Long = " + "42".toLong())
assert('4'.toString().toLong() == 4.toLong())
}
Output:
'4' as Long = 52
"4" as Long = 4
"42" as Long = 42
Is it a good practice to use char.toString().toLong() or there is a better way to convert char to Long?
Does "4" represented by chars? Why it is not equal to it char representation?
From the documentation:
class Char : Comparable (source) Represents a 16-bit Unicode
character. On the JVM, non-nullable values of this type are
represented as values of the primitive type char.
fun toLong(): Long
Returns the value of this character as a Long.
When you use '4' as Long you actually get the Unicode (ASCII) code of the char '4'
As mTak says, Char represents a Unicode value. If you are using Kotlin on the JVM, you can define your function as follows:
private fun Long.sumDigits() = this.toString().map(Character::getNumericValue).sum().toLong()
There's no reason to return Long rather than Int, but I've kept it the same as in your question.
Non-JVM versions of Kotlin don't have the Character class; use map {it - '0'} instead.