Question

I was interviewed yesterday and this question arise:

public class Main {

    public static void main(String[] args) {

        // Code fragment must prints true if the double variables x and y are both
        // strictly between 0 and 1 and false otherwise. 

        // Assuming args will be always correct (double) input, 
        // Where's the safety bug?

        double x,y;

        x = Double.parseDouble(args[0]);
        y = Double.parseDouble(args[1]);

        System.out.println( (0 < x && x < 1) && (0 < y && y < 1)? true : false );
    }
}

I wrote there were no bug anywhere, I mean, code works. Back at home, tried and indeed it works. But although I did get the job, this question was my only bad one, and I can’t get it out of my head (it seams fair straight simple). I believe I don’t even understand what a safety bug is so, my question then is:

What’s a safety bug and does this code indeed have one?

UPDATE:

After response I put some related resources:

Eric Lippert‘s Why does JScript have rounding errors?

Answer 1

Judging from the comment in the source:

Assuming args will be always correct (double) input, Where’s the safety bug?

There exists a case in which the program will deliver a wrong output, and there seems to be no dubious except the expression itself:

(0 < x && x < 1) && (0 < y && y < 1)

Although I haven’t tested it, there may be a problem when this expression is confronted with one of the edge cases for either x or y: NaN, +/- infinity and maybe even -0.0.

Also there exist definitively valid inputs which may violate the logical expectation of the user due to the limited precision with which number are represented in a double (e.g. x = 1E-400 is greater than 0, but parses as 0, so the expression delivers false although the user would expect true).