Help - Unexpected result

Question

I have attached a small code showing my problem.
Please help me to understand and resolve.
https://code.sololearn.com/cFQ4UJHi37mW/?ref=app

Answer 1

The explicit casting of the char type to an integer will result to their values in the ASCII Table. To get your result right, you need to cast "I" back to string then from string to an integer. This could be
total += i.toString().toInt()

Answer 2

Yes. There would be a couple JVM bytecodes for Char to String plus a loop for String to Int (var sum=0;for (i in "$c"){sum*=10; sum+=i-'0'}) for the first versus three JVM bytecodes (Byte to Int twice Int+Int) for the second.

Answer 3

Actually I don't know kotlin but I noticed that when you loop through the string and take one character, i is now type char and if you convert it to int it gonna give you asci of the char
So I did this
total += i.toString().toInt();
Am not sure it is the perfect answer but it did work

Answer 4

Your i.toString().toInt() solution requires a lot more code to execute over the i-'0' one.

Answer 5

Ruba Kh , Mirielle , Thanks

Answer 6

This would also get 10. Char minus Char returns Int of difference. Int plus Char returns Char of sum.
for(i in num) {
total += i-'0'
}

Answer 7

John Wells Thanks.
I have played with your suggestion and it works well, though will take me a bit longer to fully understand the concept.

Answer 8

'1'-'0' = 49-48 = 1
'4'-'0' = 52-48 = 4
'5'-'0' = 53-48 = 5

Answer 9

The Char class has this extended method defined.
operator fun Char.minus(other: Char): Int =
this.toInt()-other.toInt()

Answer 10

John Wells Thanks again.
I understood that part, but I am getting my head around some of the advantages this approach provides, IE: cryptology, passwords, etc.
I think I am going to have some fun learning all this. 😁👍

Answer 11

John Wells Thanks for showing me the extended method, it helps explain the background reasoning

Answer 12

John Wells hence it would be slower?