253631d73f
To digit simplification
I found out the other day that all the ascii digits have the first four bits as one would hope them to. (Eg. char `2` ends `0b0010`). There are two bits to indicate it's in the digit range ( `0b0011_0000`). If it is a true digit then all the higher bits aside from these two will be 0 (as ascii is the lowest part of the unicode u32 spectrum). So XORing with `0b11_0000` should mean we either get the number 0-9 or alternativly we get a larger number in the u32 space. If we get something that's not 0-9 then it will be discarded as it will be greater than the radix.
The code seems so fast though that there's quite a lot of noise in the benchmarks so it's not that easy to prove conclusively that it's faster as well as less instructions.
The non-fast path I was toying with as well wondering if we could do this as then we'd only have one return and less instructions still:
```
match self {
'a'..='z' => self as u32 - 'a' as u32 + 10,
'A'..='Z' => self as u32 - 'A' as u32 + 10,
_ => { radix = 10; self as u32 ^ ASCII_DIGIT_MASK},
}
```
Here's the [godbolt](https://godbolt.org/z/883c9n).
( H/T to ``@byteshadow`` for pointing out xor was what I needed)
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