rust - 如何从具有整数和分数的任意拆分的原始整数类型解析二进制补码定点数？

Question

我正在解析 OpenType 字体文件，需要解析(和写入)两种定点数:

• 16 位有符号固定数，低 14 位为小数 (2.14)
• 32 位有符号定点数 (16.16)

我假设，最后，它应该被转换为 f32

---

OpenType Spec描述:

The F2DOT14 format consists of a signed, 2’s complement integer and an unsigned fraction. To compute the actual value, take the integer and add the fraction.

2.14 值的示例是:

Decimal Value   Hex Value   Integer     Fraction
1.999939        0x7fff      1           16383/16384
1.75            0x7000      1           12288/16384
0.000061        0x0001      0           1/16384
0.0             0x0000      0           0/16384
-0.000061       0xffff      -1          16383/16384
-2.0            0x8000      -2          0/16384

---

我有一个有效的解决方案，但仅适用于 2.14 值:

fn from(number: u16) -> f32 {
    let mut int = (number >> 14) as f32;
    if int > 1f32 {
        int -= 4f32;
    }
    let frac = (number & 0b11_1111_1111_1111) as f32 / 16384 as f32;
    int + frac
}

因为整数值应该是[-2, 2), 如果解析出来的整数大于1我减去4得到负数。

我正在寻找一种方法来对定点数进行任何可能的拆分(例如 2.14、16.16、3.5 , 24.40 等)在 Rust 整数基本类型的标准范围内(u16、u32、u64、等)。

Answer 1

能够解决我的问题，这里是解析 16 位定点数的示例:

use std::mem::size_of;

fn from_u16(raw: u16, frac_count: usize) -> f32 {
  let bit_count = size_of::<u16>() * 8;
  let int_count = bit_count - frac_count;

  let unsigned = (raw >> frac_count) as isize;
  let sign_bit = unsigned >> (int_count - 1) & 1;
  let high_bits = if sign_bit == 1 { -1 } else { 0 };
  let signed = high_bits << int_count | unsigned as isize;

  let mut mask = 0;
  for i in 0..=frac_count {
    mask = mask << i | 1;
  }

  let frac = (raw & mask) as f32 / (1 << frac_count) as f32;
  signed as f32 + frac
}