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I've been trying to figure out how to determine if a double value is -0.0d. I googled the problem for a bit and found a few different approaches, however all the threads I found were at least a couple years old and seem to no longer work.
Here's what I've tried so far:
double x, y;
x = 0;
y = -0;
println(x == y);
println(Double.compare(x, y) == 0);
println(new Double(x).equals(new Double(y)));
println(Double.doubleToLongBits(x) == Double.doubleToLongBits(y));
println(Math.signum(x) == Math.signum(y));
All of those print true, unfortunately.
The reason I need to distinguish between 0 and -0, if you're wondering, is because I'm trying to parse a double value from user input, and using -0 as a sentinel value in case of an exception.
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If number<0 the number is negative. If a number is neither positive nor negative, the number is equal to 0. Let's implement the above logic in a Java program using the if-else statement.
If a number is less than zero, it is a negative number. If a number equals to zero, it is zero. Did you find this article helpful? Sorry about that. How can we improve it?
1 If number>0 the number is positive. 2 If number<0 the number is negative. 3 If a number is neither positive nor negative, the number is equal to 0.
Recommended: Please try your approach on {IDE} first, before moving on to the solution. The signed shift n>>31 converts every negative number into -1 and every other into 0.
Your problem is that you are using y = -0 instead of y = -0.0. 0 is an integer literal and there is no -0 int value, it just stays 0 and then is widened to +0.0d.You could also use -(double)0 to do the widening before the negation and get the same result as -0.0.
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