Arithmetic operations on float values return unexpected answer

Question

If i print the value of x for :

int a=1;
int b=6;
float x=(a/b);

The output is 0.0

But if i change the third line to float x = (float)a/(float)b;

The output is 0.1666667(which it should be)

Why the difference?

Many people do integer division in primary school and forget it. — Peter Lawrey, Apr 10 '16 at 14:05

score 1 · Accepted Answer · answered Apr 10 '16 at 13:49

a/b is calculated as the division of two integers and therefore its result is an integer, and so everything beyond the decimal floating point is truncated and you end up with 0.0.

For example: if you had:

int a = 6;
int b = 4;
float x =(a/b);

You would get that x = 1.0 because the result of the calculation is truncated, and then when you assign it to a float, the integer result value 1 gets cast to a float, i.e. 1.0.

When you calculate (float)a/(float)b, you're calculating the division of two float variables and the result is float (the same would happen if only one of the variables / values was float and the other one was integer).

score 1 · Answer 2 · answered Apr 10 '16 at 13:51

That is because both of the variables are of type int and hence the division returns an int.

You'll have to cast at least one of them to float in order to force the division to return the result in float.

Here is the code snippet:

public static void main (String[] args) throws Exception {
    int a = 1;
    int b = 6;
    float x = ((float)a/b);
    System.out.println("Result: " + x);
}

Output:

Result: 0.16666667

Note: In the above example, I have casted a to float. You can also cast b to float instead of a as follows:

float x = (a/(float)b);

Arithmetic operations on float values return unexpected answer

2 Answers2