To digit simplification #82094
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Remove unused const
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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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…laumeGomez Rollup of 11 pull requests Successful merges: - rust-lang#79981 (Add 'consider using' message to overflowing_literals) - rust-lang#82094 (To digit simplification) - rust-lang#82105 (Don't fail to remove files if they are missing) - rust-lang#82136 (Fix ICE: Use delay_span_bug for mismatched subst/hir arg) - rust-lang#82169 (Document that `assert!` format arguments are evaluated lazily) - rust-lang#82174 (Replace File::create and write_all with fs::write) - rust-lang#82196 (Add caveat to Path::display() about lossiness) - rust-lang#82198 (Use internal iteration in Iterator::is_sorted_by) - rust-lang#82204 (Update books) - rust-lang#82207 (rustdoc: treat edition 2021 as unstable) - rust-lang#82231 (Add long explanation for E0543) Failed merges: r? `@ghost` `@rustbot` modify labels: rollup
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I found out the other day that all the ascii digits have the first four bits as one would hope them to. (Eg. char
2ends0b0010). 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 with0b11_0000should 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:
Here's the godbolt.
( H/T to @byteshadow for pointing out xor was what I needed)