I want to divide two Integers and get a BigDecimal back in Kotlin.
E.g. 3/6 = 0.500000.
I've tried some solutions, like:
val num = BigDecimal(3.div(6))
println("%.6f".format(num))
// The result is: 0.000000
but none of them solve my problem.
3 and 6 are both Int, and dividing one Int by another gives an Int: that's why you get back 0. To get a non-integer value you need to get the result of the division to be a non-integer value. One way to do this is convert the Int to something else before dividing it, e.g.:
val num = 3.toDouble() / 6
num will now be a Double with a value of 0.5, which you can format as a string as you wish.
You might have better luck with:
val num = 3.toBigDecimal().divide(6.toBigDecimal())
println(num)
// prints 0.5
You have to convert both numbers to BigDecimal for the method to work. This will show the exact quotient, or throw an exception if the exact quotient cannot be represented (ie a non-terminating decimal).
You can set the scale and rounding mode as follows:
val num = 3.toBigDecimal().divide(6.toBigDecimal(), 4, RoundingMode.HALF_UP)
println(num)
// prints 0.5000
Link to reference article
Dividing Int by Int will give Int result only. To get float result , you need to convert one of the number to float.
You can use toFloat() function also.
var result = Int.toFloat() / Int
Related
Hi this is my first ever program I'm tryin to write in android studio/Kotlin and I'm not sure how to proceed.
so in my program i have a few math tasks to do and it does it fine but what I need to do now is separate part of the answer then covert it then print out both parts
for example if my answer was 1.5232 i would like to convert the decimal part of the answer to a string that matches a range if its in it. the ranges I have are in the .0000 area so I would like to limit the decimal range too.
so final result would look like this
1 (whatever my string in range is)
I hope I included enough info thank you in advance.
The first part of the task is to split the number into the integer and fractional components:
val input = 1.5232
val integer = input.toInt() // rounds DOWN to nearest smaller Int
val fractional = input % 1.0 // The remainder when dividing by 1.0 is the fraction
The strategy I would use to round to the nearest fractional value given a certain precision is to multiply by that precision, and round to the nearest integer. That would give you the numerator, and the precision would be the denominator:
val denominator = 8 // fractional precision
val numerator = (fractional * denominator).roundToInt() // rounds up or down to nearest Int
Then to put it together, you can use a string template:
val result = "$integer $numerator/$denominator"
println(result)
Simplifying the fraction would be another task if you need that. You can find various algorithms for finding greatest common divisor of two numbers. Use one of those and divide the numerator and denominator by that value.
I use remainder inside my code with kotlin in android project but with this value I don't get the correct answer.
variable is :
val vv = 1529.71
val ratio = 0.01
val remainder = vv.rem(ratio)
it's must be zero but remainder value is :
4.5363018896793506E-15
I don't understand why this happened.
The answer is because vv isn't actually 1529.71 but the closest possible Double, the exact value is 1529.7100000000000363797880709171295166015625 (the easiest way to see it is println(java.math.BigDecimal(vv))). If you want to represent decimal numbers exactly, use BigDecimal and pass the fraction as a string:
val vv = BigDecimal("1529.71")
val ratio = BigDecimal("0.01")
val remainder = vv.rem(ratio)
Read more about floating point here: https://floating-point-gui.de/
In my case, i had to get only the exact digits of two numbers after the decimal point.
I achieved it by doing this:
val input = 30.47f
val remainder = (input * 100).toInt() - (input.toInt() * 100)
// remainder = 47 exactly, and not 469999999...
Hope this would be helpful for someone.
I'm Cesare from Italy (please excuse my english), this is my first question posted on StackOverflow and I'm pretty new to Objective-C... I hope I won't make a mess on my first try.
I would like to "combine" two integers that I already have to create a new float (or a double).
By "combine", I mean that I'd like to have the first int before the point and the second int after the point, I'm not trying to convert from int to float. Maybe an example could explain better what I'm trying to do:
First int: 7
Second int: 92
The float I'm trying to get: 7.92
I looked for a previous question like mine but I haven't found anything, maybe because what I'm trying to do is pretty dumb (I have a UIPickerView with 2 components, each containing hundreds of integers, and I'm trying to create a float or double variable that has the selection of the first component before the point and the selection of the second component after the point).
Thanks in advance for your help,
Cesare
Just think about what the definition and/or the purpose of the decimal point is. It separates the part of the number which is less than one from the part greater than or equal to one.
So, keep dividing the part after the decimal point until it's less than 1:
int firstPart = 7;
int secondPart = 92; // or whatever
float f = secondPart;
while (f >= 1) {
f /= 10;
}
f += firstPart;
I know this is later, but came across a similar situation. Maybe this is more efficient.
Take the second number, 92 and divide it by 100. That gives you .92. Add that to the first number. That can give you 7.92. However, since you're adding integers that you want converted to a float, you'll need to cast the numbers when adding them. Like this:
int firstPart = 7;
int secondPart = 92;
float afterDecimalPlace = (float)secondPart/100.0;
float numberAsFloat = (float)firstPart + afterDecimalPlace;
essentially that is:
92/100 = .92
7 + .92 = 7.92
I want to convert a decimal number to an int without rounding it, for example:
if the number is 3.9 it will be turned into 3 (if it would have been rounded it would be 4).
You generally don't need to do anything special, as by default a cast from float/double to an integer type results in truncation:
float f = 3.9f;
int i = (int)f; // i = 3
It depends on how you want the negative values be treated. Typecasting to int would just truncate in that way that the part left of the decimal point will remain. -3.9f would turn into -3. Using floor before casting would ensure that it results in -4.
(all within the variable type boundaries of course)
you can do like this bellow :-
float myFloat = 3.9;
int result1 = (int)ceilf(myFloat );
NSLog(#"%d",result1);
int result2 = (int)roundf(myFloat );
NSLog(#"%d",result2);
int result3 = (int)floor(myFloat);
NSLog(#"%d",result3);
int result4 = (int) (myFloat);
NSLog(#"%d",result4);
OUTPUT IS
4
4
3
3
Just cast the float to an int and it will truncate your result.
You can simple typecast to int
I am working on an app that needs to utilize a ratio of a given number and multiply that ratio times another number. Problem is that I can't get numbers less that 1 to give me the proper decimal ratio, instead it gives me zero (when it should be .5).
Example:
float number = 1/2; // This gives me zero
double number = 1/2; // This also gives me zero
If you don't specify decimal places you're using integers which means the calculation is performed with integer precision before the result is cast to the type on the LHS. You want to do the the following when using hard coded numbers in your code:
float number = 1.0f / 2.0f;
double number = 1.0 / 2.0;
If you're aiming to use integer variables for an operation, you'll want to cast them to the type that you want for your result.
Try this
float number = 1.0/2.0;
Remember that 1 is an int, so you are essentially taking
(int)1 / (int)2
which returns
(int)0
To cast variables that are ints, do
float number = (float)numerator / (float)denominator;