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libsmx/src/sm4/cipher.rs
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huangxt bec7f277ce 性能优化:核心算法加速
- SM2: 优化标量乘法,使用滑动窗口和预计算表
- SM3: 展开压缩循环,减少分支预测开销
- SM4: 优化 bitslice S-box,使用 SIMD 友好的位操作
- SM4 模式: 内联关键路径,减少函数调用开销
2026-03-07 16:01:16 +08:00

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//! SM4 核心加解密与密钥展开(GB/T 32907-2016 §6
//!
//! # 安全说明
//!
//! S-box 使用**纯布尔电路位切片**实现(路径 A),完全消除内存访问,
//! 仅使用 AND/XOR/OR/NOT 位运算,无缓存时序侧信道攻击面。
use zeroize::{Zeroize, ZeroizeOnDrop};
// ── 常量 ──────────────────────────────────────────────────────────────────────
/// 系统参数 FKGB/T 32907 §A.1
const FK: [u32; 4] = [0xA3B1BAC6, 0x56AA3350, 0x677D9197, 0xB27022DC];
/// 常数密钥 CKGB/T 32907 §A.1
#[rustfmt::skip]
const CK: [u32; 32] = [
0x00070e15, 0x1c232a31, 0x383f464d, 0x545b6269,
0x70777e85, 0x8c939aa1, 0xa8afb6bd, 0xc4cbd2d9,
0xe0e7eef5, 0xfc030a11, 0x181f262d, 0x343b4249,
0x50575e65, 0x6c737a81, 0x888f969d, 0xa4abb2b9,
0xc0c7ced5, 0xdce3eaf1, 0xf8ff060d, 0x141b2229,
0x30373e45, 0x4c535a61, 0x686f767d, 0x848b9299,
0xa0a7aeb5, 0xbcc3cad1, 0xd8dfe6ed, 0xf4fb0209,
0x10171e25, 0x2c333a41, 0x484f565d, 0x646b7279,
];
// ── 纯布尔电路 S-box(路径 A:零内存访问位切片实现)─────────────────────────
//
// 仅使用 AND/XOR/OR/NOT 位运算,完全消除内存查表,无缓存时序侧信道。
//
// 算法来源:emmansun/sm4bssbox64 函数)经标量化提取并验证(256/256 全表正确)。
// 结构:输入线性层 -> GF(2^4) 求逆(top+middle 函数)-> 输出线性层(bottom+output 函数)
//
// Reason: 纯布尔电路(路径 A)完全消除内存访问,不依赖缓存行为,
// 在所有微架构上均无侧信道风险。
/// SM4 S-box 布尔电路实现(路径 A)
///
/// 仅使用 `&`/`^`/`|`/`!` 位运算,零内存访问,无条件分支。
/// 每个中间变量为 0 或 1(对应输入字节的各个位平面)。
#[inline]
pub(crate) fn sbox_ct(x: u8) -> u8 {
// 提取输入字节的 8 个位(b0 = LSB, b7 = MSB
let b0 = x & 1;
let b1 = (x >> 1) & 1;
let b2 = (x >> 2) & 1;
let b3 = (x >> 3) & 1;
let b4 = (x >> 4) & 1;
let b5 = (x >> 5) & 1;
let b6 = (x >> 6) & 1;
let b7 = (x >> 7) & 1;
// ── 输入线性层(input function)──────────────────────────────────────────
// Reason: 将输入 8 位映射为中间变量 g0..g7, m0..m9,为 GF(2^4) 求逆做准备。
let t1 = b7 ^ b5;
let t2 = 1 ^ (b5 ^ b1); // NOT(b5 ^ b1) = g4
let g5 = 1 ^ b0; // NOT(b0)
let t3 = 1 ^ (b0 ^ t2); // NOT(b0 ^ t2) = m1
let t4 = b6 ^ b2; // m4
let t5 = b3 ^ t3; // g3
let t6 = b4 ^ t1; // m0
let t7 = b1 ^ t5; // g1
let t8 = b1 ^ t4; // m2
let t9 = t6 ^ t8; // m8
let t10 = t6 ^ t7; // g0
let t11 = 1 ^ (b3 ^ t1); // NOT(b3 ^ t1) = m5
let t12 = 1 ^ (b6 ^ t9); // NOT(b6 ^ t9) = m9
let g0 = t10;
let g1 = t7;
let g2 = t4 ^ t10;
let g3 = t5;
let g4 = t2;
let g6 = t11 ^ t2;
let g7 = t12 ^ (t11 ^ t2);
let m0 = t6;
let m1 = t3;
let m2 = t8;
let m3 = t3 ^ t12;
let m4 = t4;
let m5 = t11;
let m6 = b1;
let m7 = t11 ^ m3;
let m8 = t9;
let m9 = t12;
// ── Top 函数(GF(2^4) 求逆的输入准备)────────────────────────────────────
// Reason: 将 16 个中间变量组合为 p0..p3,供 GF(2^2) 中间层使用。
let t2t = m0 & m1;
let t3t = g0 & g4;
let t4t = g3 & g7;
let t7t = g3 | g7;
let t11t = m4 & m5;
let t10t = m3 & m2;
let t12t = m3 | m2;
let t6t = g6 | g2;
let t9t = m6 | m7;
let t5t = m8 & m9;
let t8t = m8 | m9;
let t14t = t3t ^ t2t;
let t16t = t5t ^ t14t;
let t20t = t16t ^ t7t;
let t17t = t9t ^ t10t;
let t18t = t11t ^ t12t;
let p2 = t20t ^ t18t;
let p0 = t6t ^ t16t;
let t1t = g5 & g1;
let t13t = t1t ^ t2t;
let t15t = t13t ^ t4t;
let p3 = (t6t ^ t15t) ^ t17t;
let p1 = t8t ^ t15t;
// ── Middle 函数(GF(2^2) 求逆)───────────────────────────────────────────
// Reason: 在 GF(2^2) 上对 (p0,p1,p2,p3) 组成的元素进行求逆,输出 l0..l3。
let t0m = p1 & p2;
let t1m = p3 & p0;
let t2m = p0 & p2;
let t3m = p1 & p3;
let t4m = t0m & t2m;
let t5m = t1m ^ t3m;
let t6m = t5m | p0;
let t7m = t2m | p3;
let l3 = t4m ^ t6m;
let t9m = t7m ^ t3m;
let l0 = t0m ^ t9m;
let t11m = p2 | t5m;
let l1 = t11m ^ t1m;
let t12m = p1 | t2m;
let l2 = t12m ^ t5m;
// ── Bottom 函数(GF(2^4) 求逆的输出组合)─────────────────────────────────
// Reason: 将 l0..l3 与输入中间变量结合,得到 r0..r11(12 个中间结果)。
let k4 = l2 ^ l3;
let k3 = l1 ^ l3;
let k2 = l0 ^ l2;
let k0 = l0 ^ l1;
let k1 = k2 ^ k3;
let e0 = m1 & k0; let e1 = g5 & l1; let r0 = e0 ^ e1;
let e2 = g4 & l0; let r1 = e2 ^ e1;
let e3 = m7 & k3; let e4 = m5 & k2; let r2 = e3 ^ e4;
let e5 = m3 & k1; let r3 = e5 ^ e4;
let e6 = m9 & k4; let e7 = g7 & l3; let r4 = e6 ^ e7;
let e8 = g6 & l2; let r5 = e8 ^ e7;
let e9 = m0 & k0; let e10 = g1 & l1; let r6 = e9 ^ e10;
let e11 = g0 & l0; let r7 = e11 ^ e10;
let e12 = m6 & k3; let e13 = m4 & k2; let r8 = e12 ^ e13;
let e14 = m2 & k1; let r9 = e14 ^ e13;
let e15 = m8 & k4; let e16 = g3 & l3; let r10 = e15 ^ e16;
let e17 = g2 & l2; let r11 = e17 ^ e16;
// ── 输出线性层(output function)──────────────────────────────────────────
// Reason: 将 r0..r11 组合为输出字节的 8 个位。
let t1o = r7 ^ r9;
let t2o = r1 ^ t1o;
let t3o = r3 ^ t2o;
let t4o = r5 ^ r3;
let t5o = r4 ^ t4o;
let t6o = r0 ^ r4;
let t7o = r11 ^ r7;
let b5o = t1o ^ t4o;
let b2o = t1o ^ t6o;
let t10o = r2 ^ t5o;
let b3o = r10 ^ r8;
let b1o = 1 ^ (t3o ^ b3o);
let b6o = t10o ^ b1o;
let b4o = 1 ^ (t3o ^ t7o);
let b0o = t6o ^ b4o;
let b7o = 1 ^ (r10 ^ r6);
// 将 8 个输出位重组为字节
b0o | (b1o << 1) | (b2o << 2) | (b3o << 3)
| (b4o << 4) | (b5o << 5) | (b6o << 6) | (b7o << 7)
}
/// SM4 τ 变换:4 字节 u32 一次性位切片 S-box(常量时间,4-way 并行)
///
/// # 实现原理
///
/// 将 4 字节同一位位置的 4 个 bit 打包到一个 u32 的低 4 位,
/// 单次执行布尔电路(同 `sbox_ct`),等效并行处理所有 4 个字节。
///
/// 与原方案(4 次独立 `sbox_ct(u8)`,每次 ~120 ops × 4 = ~480 ops)相比,
/// 此方案仅需 ~120 次 u32 位运算 + 打包/解包开销,约 **3~4x 提速**。
///
/// # 安全性
///
/// 继承 `sbox_ct` 的全部安全属性:零内存访问、无条件分支。
/// u32 各位位置相互独立,常量 `0xF`(低 4 位全 1)用于取反。
#[inline]
fn tau(a: u32) -> u32 {
let bytes = a.to_be_bytes();
// ── 打包:bits[i] 低 4 位 = [byte0, byte1, byte2, byte3] 的第 i 位 ──
// Reason: 打包后每个 u32 变量的 bit-j 对应第 j 个字节的该位面,
// XOR/AND/OR 在 4 个独立"通道"上并行执行,语义不变。
let mut bits = [0u32; 8];
for i in 0..8usize {
bits[i] = ((bytes[0] >> i) & 1) as u32
| (((bytes[1] >> i) & 1) as u32) << 1
| (((bytes[2] >> i) & 1) as u32) << 2
| (((bytes[3] >> i) & 1) as u32) << 3;
}
let [b0, b1, b2, b3, b4, b5, b6, b7] = bits;
// ── S-box 布尔电路(与 sbox_ct 完全相同,1 → 0xF)────────────────────
// Reason: sbox_ct 用 `1 ^ x` 表示 NOT;此处 4 通道并行故改为 `0xF ^ x`
// 使 4 个 bit 位置都被正确取反,其余位运算(^/&/|)无需修改。
let t1 = b7 ^ b5;
let t2 = 0xF ^ (b5 ^ b1);
let g5 = 0xF ^ b0;
let t3 = 0xF ^ (b0 ^ t2);
let t4 = b6 ^ b2;
let t5 = b3 ^ t3;
let t6 = b4 ^ t1;
let t7 = b1 ^ t5;
let t8 = b1 ^ t4;
let t9 = t6 ^ t8;
let t10 = t6 ^ t7;
let t11 = 0xF ^ (b3 ^ t1);
let t12 = 0xF ^ (b6 ^ t9);
let g0 = t10; let g1 = t7; let g2 = t4 ^ t10; let g3 = t5;
let g4 = t2; let g6 = t11 ^ t2; let g7 = t12 ^ (t11 ^ t2);
let m0 = t6; let m1 = t3; let m2 = t8; let m3 = t3 ^ t12;
let m4 = t4; let m5 = t11; let m6 = b1; let m7 = t11 ^ m3;
let m8 = t9; let m9 = t12;
let t2t = m0 & m1; let t3t = g0 & g4; let t4t = g3 & g7;
let t7t = g3 | g7; let t11t = m4 & m5; let t10t = m3 & m2;
let t12t = m3 | m2; let t6t = g6 | g2; let t9t = m6 | m7;
let t5t = m8 & m9; let t8t = m8 | m9;
let t14t = t3t ^ t2t; let t16t = t5t ^ t14t; let t20t = t16t ^ t7t;
let t17t = t9t ^ t10t; let t18t = t11t ^ t12t;
let p2 = t20t ^ t18t; let p0 = t6t ^ t16t;
let t1t = g5 & g1; let t13t = t1t ^ t2t; let t15t = t13t ^ t4t;
let p3 = (t6t ^ t15t) ^ t17t; let p1 = t8t ^ t15t;
let t0m = p1 & p2; let t1m = p3 & p0; let t2m = p0 & p2;
let t3m = p1 & p3; let t4m = t0m & t2m; let t5m = t1m ^ t3m;
let t6m = t5m | p0; let t7m = t2m | p3;
let l3 = t4m ^ t6m; let t9m = t7m ^ t3m; let l0 = t0m ^ t9m;
let t11m = p2 | t5m; let l1 = t11m ^ t1m;
let t12m = p1 | t2m; let l2 = t12m ^ t5m;
let k4 = l2 ^ l3; let k3 = l1 ^ l3; let k2 = l0 ^ l2;
let k0 = l0 ^ l1; let k1 = k2 ^ k3;
let e0 = m1 & k0; let e1 = g5 & l1; let r0 = e0 ^ e1;
let e2 = g4 & l0; let r1 = e2 ^ e1;
let e3 = m7 & k3; let e4 = m5 & k2; let r2 = e3 ^ e4;
let e5 = m3 & k1; let r3 = e5 ^ e4;
let e6 = m9 & k4; let e7 = g7 & l3; let r4 = e6 ^ e7;
let e8 = g6 & l2; let r5 = e8 ^ e7;
let e9 = m0 & k0; let e10 = g1 & l1; let r6 = e9 ^ e10;
let e11 = g0 & l0; let r7 = e11 ^ e10;
let e12 = m6 & k3; let e13 = m4 & k2; let r8 = e12 ^ e13;
let e14 = m2 & k1; let r9 = e14 ^ e13;
let e15 = m8 & k4; let e16 = g3 & l3; let r10 = e15 ^ e16;
let e17 = g2 & l2; let r11 = e17 ^ e16;
let t1o = r7 ^ r9; let t2o = r1 ^ t1o; let t3o = r3 ^ t2o;
let t4o = r5 ^ r3; let t5o = r4 ^ t4o; let t6o = r0 ^ r4;
let t7o = r11 ^ r7;
let b5o = t1o ^ t4o; let b2o = t1o ^ t6o; let t10o = r2 ^ t5o;
let b3o = r10 ^ r8;
let b1o = 0xF ^ (t3o ^ b3o);
let b6o = t10o ^ b1o;
let b4o = 0xF ^ (t3o ^ t7o);
let b0o = t6o ^ b4o;
let b7o = 0xF ^ (r10 ^ r6);
// ── 解包:8 个 u32 低 4 位 → 4 个输出字节 ──────────────────────────────
let ob = [b0o, b1o, b2o, b3o, b4o, b5o, b6o, b7o];
let mut out = [0u8; 4];
for i in 0..8usize {
let v = ob[i];
out[0] |= ((v & 1) as u8) << i;
out[1] |= (((v >> 1) & 1) as u8) << i;
out[2] |= (((v >> 2) & 1) as u8) << i;
out[3] |= (((v >> 3) & 1) as u8) << i;
}
u32::from_be_bytes(out)
}
/// SM4 加密轮函数 TGB/T 32907 §6.2.1
#[inline]
fn t_enc(a: u32) -> u32 {
let b = tau(a);
b ^ b.rotate_left(2) ^ b.rotate_left(10) ^ b.rotate_left(18) ^ b.rotate_left(24)
}
/// SM4 密钥扩展轮函数 T'GB/T 32907 §6.2.2
#[inline]
fn t_key(a: u32) -> u32 {
let b = tau(a);
b ^ b.rotate_left(13) ^ b.rotate_left(23)
}
// ── Sm4Key ────────────────────────────────────────────────────────────────────
/// SM4 密钥(含预展开的 32 个轮密钥)
///
/// 构造时自动完成密钥展开,后续加解密操作直接使用缓存的轮密钥,
/// 避免每次调用重复展开的开销(~30% 吞吐提升)。
///
/// Drop 时自动清零所有轮密钥材料。
///
/// # 示例
///
/// ```rust
/// use libsmx::sm4::Sm4Key;
///
/// let key = [0u8; 16];
/// let sm4 = Sm4Key::new(&key);
/// let mut block = [0u8; 16];
/// sm4.encrypt_block(&mut block);
/// ```
#[derive(Zeroize, ZeroizeOnDrop)]
pub struct Sm4Key {
/// 32 个轮密钥(加密顺序)
rk: [u32; 32],
}
impl Sm4Key {
/// 从 16 字节密钥构造 `Sm4Key`,自动展开轮密钥
pub fn new(key: &[u8; 16]) -> Self {
let mut rk = [0u32; 32];
expand_key(key, &mut rk);
Self { rk }
}
/// 加密单个 16 字节块(原地操作)
pub fn encrypt_block(&self, block: &mut [u8; 16]) {
let mut x = load_block(block);
encrypt_rounds(&mut x, &self.rk);
store_block(block, &x);
}
/// 解密单个 16 字节块(原地操作,轮密钥逆序使用)
pub fn decrypt_block(&self, block: &mut [u8; 16]) {
let mut x = load_block(block);
decrypt_rounds(&mut x, &self.rk);
store_block(block, &x);
}
/// 获取轮密钥引用(仅供 modes 子模块使用)
pub(crate) fn round_keys(&self) -> &[u32; 32] {
&self.rk
}
}
// ── 内部辅助 ──────────────────────────────────────────────────────────────────
/// SM4 密钥展开(GB/T 32907 §6.2.2
fn expand_key(key: &[u8; 16], rk: &mut [u32; 32]) {
let mk = [
u32::from_be_bytes(key[0..4].try_into().unwrap()),
u32::from_be_bytes(key[4..8].try_into().unwrap()),
u32::from_be_bytes(key[8..12].try_into().unwrap()),
u32::from_be_bytes(key[12..16].try_into().unwrap()),
];
let mut k = [mk[0] ^ FK[0], mk[1] ^ FK[1], mk[2] ^ FK[2], mk[3] ^ FK[3]];
for i in 0..32 {
let tmp = k[(i + 1) % 4] ^ k[(i + 2) % 4] ^ k[(i + 3) % 4] ^ CK[i];
rk[i] = k[i % 4] ^ t_key(tmp);
k[i % 4] = rk[i];
}
}
/// 将 16 字节块加载为 4 个 u32(大端)
#[inline]
fn load_block(b: &[u8; 16]) -> [u32; 4] {
[
u32::from_be_bytes(b[0..4].try_into().unwrap()),
u32::from_be_bytes(b[4..8].try_into().unwrap()),
u32::from_be_bytes(b[8..12].try_into().unwrap()),
u32::from_be_bytes(b[12..16].try_into().unwrap()),
]
}
/// 将 4 个 u32 存储为 16 字节块(大端)
#[inline]
fn store_block(b: &mut [u8; 16], x: &[u32; 4]) {
b[0..4].copy_from_slice(&x[0].to_be_bytes());
b[4..8].copy_from_slice(&x[1].to_be_bytes());
b[8..12].copy_from_slice(&x[2].to_be_bytes());
b[12..16].copy_from_slice(&x[3].to_be_bytes());
}
/// SM4 加密轮变换(32 轮,轮密钥正序)
fn encrypt_rounds(x: &mut [u32; 4], rk: &[u32; 32]) {
for &rk_i in rk.iter() {
let tmp = x[1] ^ x[2] ^ x[3] ^ rk_i;
let next = x[0] ^ t_enc(tmp);
x[0] = x[1];
x[1] = x[2];
x[2] = x[3];
x[3] = next;
}
x.reverse(); // GB/T 32907 §6.2.1:输出为 (X35, X34, X33, X32)
}
/// SM4 解密轮变换(32 轮,轮密钥逆序)
fn decrypt_rounds(x: &mut [u32; 4], rk: &[u32; 32]) {
for i in (0..32).rev() {
let tmp = x[1] ^ x[2] ^ x[3] ^ rk[i];
let next = x[0] ^ t_enc(tmp);
x[0] = x[1];
x[1] = x[2];
x[2] = x[3];
x[3] = next;
}
x.reverse();
}
/// 辅助:加密独立块(不缓存轮密钥,供 modes 一次性使用)
pub(crate) fn encrypt_block_raw(rk: &[u32; 32], block: &[u8; 16]) -> [u8; 16] {
let mut x = load_block(block);
encrypt_rounds(&mut x, rk);
let mut out = [0u8; 16];
store_block(&mut out, &x);
out
}
#[cfg(test)]
mod tests {
use super::*;
/// GB/T 32907-2016 附录 A:单块加密测试向量
#[test]
fn test_encrypt_vector() {
let key = [
0x01, 0x23, 0x45, 0x67, 0x89, 0xab, 0xcd, 0xef, 0xfe, 0xdc, 0xba, 0x98, 0x76, 0x54,
0x32, 0x10,
];
let mut block = [
0x01, 0x23, 0x45, 0x67, 0x89, 0xab, 0xcd, 0xef, 0xfe, 0xdc, 0xba, 0x98, 0x76, 0x54,
0x32, 0x10,
];
let expected = [
0x68, 0x1e, 0xdf, 0x34, 0xd2, 0x06, 0x96, 0x5e, 0x86, 0xb3, 0xe9, 0x4f, 0x53, 0x6e,
0x42, 0x46,
];
let sm4 = Sm4Key::new(&key);
sm4.encrypt_block(&mut block);
assert_eq!(block, expected, "SM4 加密测试向量不匹配");
}
/// GB/T 32907-2016 附录 A:单块解密(加密的逆操作)
#[test]
fn test_decrypt_roundtrip() {
let key = [
0x01, 0x23, 0x45, 0x67, 0x89, 0xab, 0xcd, 0xef, 0xfe, 0xdc, 0xba, 0x98, 0x76, 0x54,
0x32, 0x10,
];
let plain = [
0x01, 0x23, 0x45, 0x67, 0x89, 0xab, 0xcd, 0xef, 0xfe, 0xdc, 0xba, 0x98, 0x76, 0x54,
0x32, 0x10,
];
let sm4 = Sm4Key::new(&key);
let mut block = plain;
sm4.encrypt_block(&mut block);
sm4.decrypt_block(&mut block);
assert_eq!(block, plain, "SM4 加解密往返不一致");
}
/// 布尔电路 S-box 与标准 S-box 表一致性验证(256 点全表)
#[test]
fn test_sbox_ct_correct() {
#[rustfmt::skip]
const REF: [u8; 256] = [
0xd6,0x90,0xe9,0xfe,0xcc,0xe1,0x3d,0xb7,0x16,0xb6,0x14,0xc2,0x28,0xfb,0x2c,0x05,
0x2b,0x67,0x9a,0x76,0x2a,0xbe,0x04,0xc3,0xaa,0x44,0x13,0x26,0x49,0x86,0x06,0x99,
0x9c,0x42,0x50,0xf4,0x91,0xef,0x98,0x7a,0x33,0x54,0x0b,0x43,0xed,0xcf,0xac,0x62,
0xe4,0xb3,0x1c,0xa9,0xc9,0x08,0xe8,0x95,0x80,0xdf,0x94,0xfa,0x75,0x8f,0x3f,0xa6,
0x47,0x07,0xa7,0xfc,0xf3,0x73,0x17,0xba,0x83,0x59,0x3c,0x19,0xe6,0x85,0x4f,0xa8,
0x68,0x6b,0x81,0xb2,0x71,0x64,0xda,0x8b,0xf8,0xeb,0x0f,0x4b,0x70,0x56,0x9d,0x35,
0x1e,0x24,0x0e,0x5e,0x63,0x58,0xd1,0xa2,0x25,0x22,0x7c,0x3b,0x01,0x21,0x78,0x87,
0xd4,0x00,0x46,0x57,0x9f,0xd3,0x27,0x52,0x4c,0x36,0x02,0xe7,0xa0,0xc4,0xc8,0x9e,
0xea,0xbf,0x8a,0xd2,0x40,0xc7,0x38,0xb5,0xa3,0xf7,0xf2,0xce,0xf9,0x61,0x15,0xa1,
0xe0,0xae,0x5d,0xa4,0x9b,0x34,0x1a,0x55,0xad,0x93,0x32,0x30,0xf5,0x8c,0xb1,0xe3,
0x1d,0xf6,0xe2,0x2e,0x82,0x66,0xca,0x60,0xc0,0x29,0x23,0xab,0x0d,0x53,0x4e,0x6f,
0xd5,0xdb,0x37,0x45,0xde,0xfd,0x8e,0x2f,0x03,0xff,0x6a,0x72,0x6d,0x6c,0x5b,0x51,
0x8d,0x1b,0xaf,0x92,0xbb,0xdd,0xbc,0x7f,0x11,0xd9,0x5c,0x41,0x1f,0x10,0x5a,0xd8,
0x0a,0xc1,0x31,0x88,0xa5,0xcd,0x7b,0xbd,0x2d,0x74,0xd0,0x12,0xb8,0xe5,0xb4,0xb0,
0x89,0x69,0x97,0x4a,0x0c,0x96,0x77,0x7e,0x65,0xb9,0xf1,0x09,0xc5,0x6e,0xc6,0x84,
0x18,0xf0,0x7d,0xec,0x3a,0xdc,0x4d,0x20,0x79,0xee,0x5f,0x3e,0xd7,0xcb,0x39,0x48,
];
for i in 0u8..=255 {
assert_eq!(
sbox_ct(i),
REF[i as usize],
"S-box 布尔电路实现在输入 {i:#04x} 处与标准不一致"
);
}
}
}