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Given an array of Long values in Kotlin and a position parameter, when the position is in the middle of two of the array values, return the lowest

Spotify's lyrics API provides an Array of miliseconds to mark when the lyric line has changed. Having a Media Player that updates it's position every 50ms, how should i code in Kotlin the way to find the correct lyric line? The position param can be in the middle of two values of the array, so I want to get the lowest one of that two.
I tried to get the lowest value compared to the position parameter but lol, it will always be the first value of the Array... Silly fault of mine.

The problem is that I have a third one that indicates the range of that two value. For example: I have an Array of [45, 78, 125, 198]. If I pass the position param where it's value is 95, I want to return the 78 (that is the lowest value from itself, the position param and 125).
/** Input data for example (from your comment). */
val yourArray = arrayOf(45, 78, 125, 198)
val inputValue = 95
/** How to get needed index. */
val resultIndex = yourArray.indexOfLast { it < inputValue }.takeIf { it != -1 }
If you get resultIndex == null - it means you don't have value inside your array which lower then your inputValue.
I think it's simpler than find indexOfFirst and compare result later. And absolutely better and safer when sort() solution.

Insert the position param into the array, sort it, find its index and use it to get the closest value.
val array: MutableList<Long> = mutableListOf(4L, 9L, 5L, 1L)
val position = 7L
array.add(position)
println(array[array.sorted().indexOf(position) - 1])
Output: 5

If I correctly understand, you need simply use min function for compare two numbers in Kotlin and find the low one: link

Here’s a way to do it without having to make two copies of the list and doing a sort.
val foundIndex = lyricTimesList.indexOfFirst { it > inputTime }
val result = if (foundIndex == -1) lyricTimesList.size - 1 else foundIndex - 1
Note the result could be -1 if the input time is less than the first number in the list.

Starts comparing from the first number, not the whole number in kotlin

Hello I need to compare 2 numbers and I used >, => but it doesn't compare whole number, it looks for the leftest(left) number and compare
for example the number is 92,236 and i want to compare it with 100,000, it says 92236 is bigger than 100,000 and it is because of the first number which is 9 and the first number of second number that is 1 so it says 100,000 is not bigger than 9236
here what I had done
class IncreaseMoneyFragment : Fragment() {
var decide = ""
val increaseEditText = mIncreaseMoneyBinding.increaseEdt.text.toString() (get value of edit text)
val currentPayment = it.payment (get loanPayment from database)
if (increaseEditText > currentPayment) {
Toast.makeText(activity, "more", Toast.LENGTH_SHORT).show()
val more = "بیشتر"
decide = more
} else {
Toast.makeText(activity, "less", Toast.LENGTH_SHORT).show()
val less = "کمتر"
decide = less
}
builder.setTitle(" مبلغ مورد نظر از مبلغ قسط وام $decide است. ادامه میدهید؟")
THANKS FOR HELPING ME :)

You are most likely comparing strings (text) and not numbers here. That's why it's using the alphabetical order instead of the integer order:
println("92236" > "100000") // true
println(92236 > 100000) // false
You probably want to convert your strings into integers instead:
if (increaseEditText.toInt() > currentPayment.toInt()) {
// ...
}
Note that toInt will crash if the strings are not actual numbers (for instance empty).
You can use toIntOrNull if you want more safety. It returns null if the string is not a number, so you can simply check for null and deal with this problem separately before comparing.

How to count the number of string characters in Kotlin?

#SuppressLint("SetTextI18n")
private fun countName(x: String) {
val textView: TextView = findViewById(R.id.result)
for (i in 0 until x.length) {
textView.text = i.toString()
}
}
This is what I have so far. It works somewhat but for some reason it is lagging behind 1 number. For example, if I type James, the number it will output would be 4.

How to get the number of characters in a string (usually what you want):
str.length
How to get the number of Unicode code points in a string (because sometimes a code point is encoded as multiple characters):
str.codePointCount(0, str.length)
See this Java answer for more details on the difference between these two options.

There are many simple ways to get the number of characters in a string. but if you want to know the mistake in your code you should have started to count from 1 instead of 0.
for (i in 1 until x.length) {
textView.text = i.toString()
}

Kotlin float number with 2 decimals to string without precision loss

In Android-Kotlin I am getting float number from backend (for example num = 10000000.47)
When I try to String.format it and add that number in my balanceTextview it shows it with exponent (something like 1.0E10).
I want to show number normally without exponent and with 2 decimals. (Without presicion loss!)
Tried to use DecimalFormat("#.##") but it didn't help me. Maybe I'm doing something wrong?
num = 10000000.47f
val dec = DecimalFormat("#.##")
var result = dec.format(num)
my result is: 10000000
It losts my decimal places

The issue is your number type. According to the documentation:
For variables initialized with fractional numbers, the compiler infers the Double type. To explicitly specify the Float type for a value, add the suffix f or F. If such a value contains more than 6-7 decimal digits, it will be rounded.
With an example that shows how information may get lost:
val pi = 3.14 // Double
val e = 2.7182818284 // Double
val eFloat = 2.7182818284f // Float, actual value is 2.7182817
If the value is specified as Double instead of Float, i.e.
val num = 10000000.47
instead of
val num = 10000000.47f
then your approach works as expected, but could be shortened to:
"%.2f".format(num)
(note that the shorter version will also print "100" as "100.00" which is different from your approach but potentially still desired behaviour)
If you receive a Float from the backend then the information is already lost on your side. Otherwise you should be able to fix the issue by improved parsing.

The extension function format is only available in the JVM. In Kotlin/native, you can use this instead:
fun Float.toPrecision(precision: Int) =
this.toDouble().toPrecision(precision)
fun Double.toPrecision(precision: Int) =
if (precision < 1) {
"${this.roundToInt()}"
} else {
val p = 10.0.pow(precision)
val v = (abs(this) * p).roundToInt()
val i = floor(v / p)
var f = "${floor(v - (i * p)).toInt()}"
while (f.length < precision) f = "0$f"
val s = if (this < 0) "-" else ""
"$s${i.toInt()}.$f"
}

Prevent numbers from changing according to the locale in android

When the user change the locale in device the numbers are also getting changed according to the selected locale. This is causing NumberFormatException while performing mathematical operations and app is getting crashed. The code snippet which is causing the crash is given below.
public static double ToDataUnitMB(double _dataBytes){
double dDataBytes;
dDataBytes = Double.parseDouble(getDecimalFormat().format(_dataBytes / 1048576));
return dDataBytes; }
This code snippet is causing NumberFormatException and the value in _dataBytes is shown as "७२.४१". Can anyone help me to prevent the number from changing when user change the locale.
Update
I am getting the value "७२.४१" after performing the below operation getDecimalFormat().format(_dataBytes / 1048576)
So while parsing to Double it is showing numberFormatException

Since you're starting with raw _dataBytes you have several options how to format number independent of the locale.
First Approach:
You can modify following snippet to your needs. It will give you the same output regardless of the user locale.
String patern = "###.##"; //your pattern as per need
Locale locale = new Locale("en", "US");
DecimalFormat decimalFormat = (DecimalFormat) NumberFormat.getNumberInstance(locale);
decimalFormat.applyPattern(patern);
double formatedDouble = Double.parseDouble(decimalFormat.format(_dataBytes/(1024*1024f)));
Keep in mind that this method also makes grouping and decimal separators to be fixed, so that comma and dot will alway be used as, respectively, grouping separator and decimal separator.
Second Approach:
If you do not strictly require Double you could generate formatted String with something similar to following method:
String generateFormatedFileSize(long _dataBytes) {
String formatedFileSize = "";
long bytes = _dataBytes;
short unit = 1024;
if (bytes < unit)
formatedFileSize = bytes + " B";
else {
int exp = (int) (Math.log(bytes) / Math.log(unit));
formatedFileSize = String.format("%.1f %sB", bytes / Math.pow(unit, exp), "KMGT".charAt(exp - 1));
}
return formatedFileSize;
}
This formatting will be sensitive to grouping separator and decimal separator, but otherwise insensitive to Locale.
For Local that uses "US" numbering format, this will give you following output:
12.5 KB
5.3 B
8.0 MB
And for Local using "European" numbering format:
12,5 KB
5,3 B
8,0 MB
Off course, these two methods are not exclusive and you could use some mix of these approaches at different parts of the App.