Is it safe to compare a float and an int in Java?

Question

Is it safe to compare a float and an integer like this?

private static void foo(float n) {
    if (n >= 1 || n <= 0) {
      // code 
    } 
}

According to the JLS (5.6.2. Binary Numeric Promotion) if one of the parameters is float, the other is converted to float before comparison. But as far as I understand, such a comparison will return true if the converted float is binary identical to the original float. How can we ensure it?

Answer 1

Yes, your specific example is fine, since both 0 and 1 can be exactly represented as a float.

Note that this is not true in general: there are many large int values that's can't be represented exactly as a float. For example, the following prints out true (Ideone) even though 2_000_000_001 does not equal 2_000_000_000:

import java.util.*;
import java.lang.*;
import java.io.*;

class FloatTest {

    private static boolean isTwoBillion(float f) {
        return f == 2_000_000_000;
    }

    public static void main (String[] args) {
        System.out.println(isTwoBillion(2_000_000_001));
    }

}

Note that, unlike float, double has wide enough mantissa to store every 32-bit int value.

However, there are long values that cannot be represented as doubles. This starts occurring at long value 9_223_372_036_854_764 which is Long.MAX_VALUE/1000 - 10. The number one greater (...765) has no double counterpart, while the following number (...766) does. Put another way, starting with ...764 as a double, incrementing the last bit of the mantissa gives ...766 when converted back to long.