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开发者_如何学运维 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?
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;
// The integer I want to convert
int myInt = 100;
// Casting of integer to float
float newFloat = (float) myInt
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.
You just need to transfer the first value to float, before it gets involved in further computations:
float z = x * 1.0 / y;
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 !
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.
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