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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?

mskfisher
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Roman
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    float result = ((float)x) / y? – mibollma Dec 07 '10 at 14:43
  • check this out: http://xahlee.org/java-a-day/casting.html – fasseg Dec 07 '10 at 14:45
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    possible duplicate of [Java: dividing 2 ints makes an int?](http://stackoverflow.com/questions/787700/java-dividing-2-ints-makes-an-int) – dogbane Dec 07 '10 at 14:54
  • I Would suggest you use BigDecimal because floating point (Double, Float) representations and calculations are inexact, leading to erroneous results. – Vinay Lodha Dec 07 '10 at 15:05
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    @VinAy: I would **not** recommend that in this case. `BigDecimal` is not a universally correct replacement for `float`. – Matt Ball Dec 07 '10 at 15:09
  • @Matt: Yeah I agree.. but accuracy in Math operations is very important. Better not to write buggy program :) – Vinay Lodha Dec 07 '10 at 15:18
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    @VinAy: It depends entirely what the purpose is. If he is doing finance, you are correct. If he is computing asteroid trajectories, you aren't. – user207421 Dec 07 '10 at 22:45
  • I usually just multiply by 1.0, is that a bad idea? Like float z = x*1.0/y; – medopal Feb 23 '12 at 07:47

6 Answers6

169

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;
Matt Ball
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  • Is this more efficient, than: float z = (1.0 * x) / y; ? Is float conversion internally more efficient than multiplication? Tnx! – MSquare Sep 13 '13 at 09:34
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    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.](http://stackoverflow.com/q/504103/139010) – Matt Ball Sep 13 '13 at 14:15
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    The first and second one will cause errors on certain arm devices, make sure you cast both integers. – Oliver Dixon Jul 26 '14 at 01:00
  • java.lang.Integer cannot be cast to java.lang.Float – a.s.p. Oct 06 '14 at 11:48
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    @user3002853 read about boxed types vs primitives. http://docs.oracle.com/javase/tutorial/java/data/autoboxing.html – Matt Ball Oct 06 '14 at 12:28
7

You just need to transfer the first value to float, before it gets involved in further computations:

float z = x * 1.0 / y;
user unknown
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6

// The integer I want to convert

int myInt = 100;

// Casting of integer to float

float newFloat = (float) myInt
tej shah
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6

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.

Peter Lawrey
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3

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 !

user229044
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LaGrandMere
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1

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.

LostOblivion
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  • 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. – user unknown Apr 28 '12 at 00:40