Why is there a "d" in the definition of Double.NaN = 0.0d / 0.0?
java
Solution
According to the Oak language spec, the format of floating point literals were:
- 2.0d or 2.0D double
- 2.0f or 2.0F or 2.0 float
but this changed to the familiar Java way by Java version 1.0
A floating-point literal is of type float if it is suffixed with an ASCII letter F or f; otherwise its type is double and it can optionally be suffixed with an ASCII letter D or d.
The change was perhaps made to make it consistent with C-like languages, where the lack of suffix means a double.
So the `d` appears to be an historical relic; although, in the linked version of the Oak spec (which is "preliminary"), there is a margin note saying that NaN isn't implemented yet. Perhaps it was implemented in a slightly later version, and has remained the same forever after.
(Props to Mark Rotteveel for the nudge to look up the Oak language spec).
Problem
I just came across the definition of `NaN` in `Double.class`. It says: ``` /** * A constant holding a Not-a-Number (NaN) value of type * {@code double}. It is equivalent to the value returned by * {@code Double.longBitsToDouble(0x7ff8000000000000L)}. */ public static final double NaN = 0.0d / 0.0; ``` I know that according the Java specification these literals represent the same number: `0.0`, `0.0d`, and `0.0D`. Also for other constants, they did not use the 'd' suffix: ``` public static final double POSITIVE_INFINITY = 1.0 / 0.0; public static final double NEGATIVE_INFINITY = -1.0 / 0.0; ``` Why did they need to write the suffix d to the first part of 0.0 in NaN definition? Was this on purpose or by chance?