CODEWITHSUNDEEP
kotlintostringlong-integer
Dennis Vash
Dennis Vash

Reputation: 53994

Why single char and "single char String" not equal when converted to long (.toLong())

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?

Upvotes: 1

Views: 87

Answers (2)

Klitos Kyriacou
Klitos Kyriacou

Reputation: 11664

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.

Upvotes: 2

user8959091
user8959091

Reputation:

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'

Upvotes: 2

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