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Compare Double.MIN_VALUE and 0

Why doesn't this code work? In fact, this is just a small part of a larger program. It has to compare Double.MIN_VALUE

with different values, it works with all values ​​except 0.

why? Thank you!

double d = Double.MIN_VALUE;
if (0. > d) {
    System.out.println("OK");
}
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5 answers


Double.MIN_VALUE

- 4.9E-324

. And that's not less 0

. But it really isn't 0

.

If you print

System.out.println(4.9E-324d > 0.);//this is true


In this sense

0.0000000000 ... 0001! = 0. But it tends to 0

Similar 4.9E-324d != 0 but tends to 0

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You are comparing decimal numbers. If you say try something like:

 System.out.println(0.000000000000001d == 0.);//print false

You will receive a lie. If I read the java docs it says:



/**
 * A constant holding the smallest positive nonzero value of type
 * {@code double}, 2<sup>-1074</sup>. It is equal to the
 * hexadecimal floating-point literal
 * {@code 0x0.0000000000001P-1022} and also equal to
 * {@code Double.longBitsToDouble(0x1L)}.
 */

So it is close to zero, but not really 0.

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Double.MIN_VALUE

is actually a bad name for a constant in Java. It is not suitable for int.MIN_VALUE

. In C # it is called Double.Epsilon which IMHO suits better.

So, Double.MIN_VALUE

not the biggest negative double that exists. IMHO such a constant doesn't even exist in JAVA by default.

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Double.MIN_VALUE

is equal to a hexadecimal floating point literal 0x0.0000000000001P-1022

and also equalDouble.longBitsToDouble(0x1L).

So, your condition is not triggered!

0


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If you want to compare the value with the smallest number (the negative value with the largest value), you can look Double.NEGATIVE_INFINITY

at which is stated in the Java documentation:

"A constant that stores negative infinity of type double. It is equal to the value returned by Double.longBitsToDouble (0xfff0000000000000L).

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