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?

• java
• integer
• floating-point

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;

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;

// The integer I want to convert

int myInt = 100;

// Casting of integer to float

float newFloat = (float) myInt

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.