Question

In Java, we can convert an int to float implicitly, which may result in loss of precision as shown in the example code below.

public class Test {
    public  static void main(String [] args) {
        int intVal = 2147483647;
        System.out.println("integer value is " + intVal);
        double doubleVal = intVal;
        System.out.println("double value is " + doubleVal);
        float floatVal = intVal;
        System.out.println("float value is " + floatVal);
        }
}

The output is

integer value is 2147483647
double value is 2.147483647E9
float value is 2.14748365E9

What is the reason behind allowing implicit conversion of int to float, when there is a loss of precision?

Answer 1

You are probably wondering:

Why is this an implicit conversion when there is a loss of information? Shouldn’t this be an explicit conversion?

And you of course have a good point. But the language designers decided that if the target type has a range large enough then an implicit conversion is allowed, even though there may be a loss of precision. Note that it is the range that is important, not the precision. A float has a greater range than an int, so it is an implicit conversion.

The Java specification says the following:

A widening conversion of an int or a long value to float, or of a long value to double, may result in loss of precision – that is, the result may lose some of the least significant bits of the value. In this case, the resulting floating-point value will be a correctly rounded version of the integer value, using IEEE 754 round-to-nearest mode.