I have two integers x and y. I need to calculate x/y and as outcome I would like to get float. For example as an outcome of 3/2 I would like to have 1.5. I thought that easiest (or the only) way to do it is to convert x and y into float type. Unfortunately, I cannot find an easy way to do it. Could you please help me with that?
6 Answers
You just need to cast at least one of the operands to a float:
float z = (float) x / y;
or
float z = x / (float) y;
or (unnecessary)
float z = (float) x / (float) y;
5 Comments
MSquare
Is this more efficient, than: float z = (1.0 * x) / y; ? Is float conversion internally more efficient than multiplication? Tnx!
Matt Ball
I don't know, but I think it's irrelevant 99% or more of the time. It's not even remotely going to be a bottleneck. If you're truly that concerned, benchmark it yourself.
Oliver Dixon
The first and second one will cause errors on certain arm devices, make sure you cast both integers.
a.s.p.
java.lang.Integer cannot be cast to java.lang.Float
Matt Ball
@user3002853 read about boxed types vs primitives. docs.oracle.com/javase/tutorial/java/data/autoboxing.html
// The integer I want to convert
int myInt = 100;
// Casting of integer to float
float newFloat = (float) myInt
1 Comment
Adam R. Turner
in Android Studio at least, the compiler will complain about incompatible types trying to do this.
You shouldn't use float unless you have to. In 99% of cases, double is a better choice.
int x = 1111111111;
int y = 10000;
float f = (float) x / y;
double d = (double) x / y;
System.out.println("f= "+f);
System.out.println("d= "+d);
prints
f= 111111.12
d= 111111.1111
Following @Matt's comment.
float has very little precision (6-7 digits) and shows significant rounding error fairly easily. double has another 9 digits of accuracy. The cost of using double instead of float is notional in 99% of cases however the cost of a subtle bug due to rounding error is much higher. For this reason, many developers recommend not using floating point at all and strongly recommend BigDecimal.
However I find that double can be used in most cases provided sensible rounding is used.
In this case, int x has 32-bit precision whereas float has a 24-bit precision, even dividing by 1 could have a rounding error. double on the other hand has 53-bit of precision which is more than enough to get a reasonably accurate result.
1 Comment
Matt Ball
You should elaborate on why
double is better than float.You just need to transfer the first value to float, before it gets involved in further computations:
float z = x * 1.0 / y;
Comments
Here is how you can do it :
public static void main(String[] args) {
// TODO Auto-generated method stub
int x = 3;
int y = 2;
Float fX = new Float(x);
float res = fX.floatValue()/y;
System.out.println("res = "+res);
}
See you !
2 Comments
Matt Ball
Use of a wrapper type (
Float) is totally unnecessary for this.
user229044
Please don't use signatures or taglines in your posts.
Sameer:
float l = new Float(x/y)
will not work, as it will compute integer division of x and y first, then construct a float from it.
float result = (float) x / (float) y;
Is semantically the best candidate.
1 Comment
user unknown
Your answer should have been a comment. Sameer will receive no notification of your post. Semantically, converting both ints to float before computing the result is needless - therefore it isn't better, than tranforming just one. It gives a wrong impression, and is therefore inferior, imho.
BigDecimalis not a universally correct replacement forfloat.