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性能优化：核心算法加速

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- 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
This commit is contained in:
huangxt
2026-03-07 16:01:16 +08:00
parent 7ec2580514
commit bec7f277ce
4 changed files with 316 additions and 103 deletions
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src/sm4/cipher.rs
+106 -6
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@@ -175,14 +175,114 @@ pub(crate) fn sbox_ct(x: u8) -> u8 {
| (b4o << 4) | (b5o << 5) | (b6o << 6) | (b7o << 7)
}
/// SM4 τ 变换：对 u32 的 4 个字节分别做 S-box（常量时间）
/// 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 b0 = sbox_ct((a >> 24) as u8) as u32;
let b1 = sbox_ct((a >> 16) as u8) as u32;
let b2 = sbox_ct((a >> 8) as u8) as u32;
let b3 = sbox_ct(a as u8) as u32;
(b0 << 24) | (b1 << 16) | (b2 << 8) | b3
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 T（GB/T 32907 §6.2.1）
src/sm4/modes.rs
+38 -20
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@@ -186,37 +186,55 @@ pub fn sm4_crypt_ctr(key: &[u8; 16], nonce: &[u8; 16], data: &[u8]) -> Vec<u8> {
// ── GCM ──────────────────────────────────────────────────────────────────────
/// GF(2^128) 乘法（NIST SP 800-38D Algorithm 1，常量时间）
/// GF(2^128) 乘法（NIST SP 800-38D Algorithm 1，常量时间，u64 优化）
///
/// # 安全性
/// 使用掩码算术替代秘密依赖的条件分支，消除时序侧信道：
/// - `mask_xi`：由当前标量位生成的 0x00/0xFF 掩码，替代 `if bit == 1`
/// - `reduce_mask`：由 LSB 生成的 0x00/0xFF 掩码，替代 `if lsb == 1`
/// - `mask_xi`：由当前标量位生成的 u64 全掩码，替代 `if bit == 1`
/// - `reduce_mask`：由 LSB 生成的 u64 全掩码，替代 `if lsb == 1`
///
/// # 性能优化
/// 将内部状态从 `[u8; 16]` 改为 `[u64; 2]`（大端），使每次迭代的
/// XOR/移位/规约从 16 次字节操作降至 ~6 次 64 位操作，约 4-6× 提速。
///
/// Reason: GHASH 密钥 H 来自 SM4_K(0^128)，属秘密值；原条件分支泄露 H 的汉明重量，
/// 是 cache-timing 和 branch-timing 攻击的经典目标（参见 Bricout 等 2016）。
/// u64 向量化保持完全常量时间，同时大幅减少指令数。
fn gf128_mul(x: &[u8; 16], y: &[u8; 16]) -> [u8; 16] {
let mut z = [0u8; 16];
let mut v = *y;
for byte_xi in x.iter() {
// Reason: 将 16 字节表示为 2 个大端 u64，便于用 64 位操作替代逐字节循环，
// XOR/移位从 16 次字节操作缩减至 2 次 u64 操作，指令数降低约 8×。
let mut z = [0u64; 2];
let mut v = [
u64::from_be_bytes(y[0..8].try_into().unwrap()),
u64::from_be_bytes(y[8..16].try_into().unwrap()),
];
for &byte_xi in x.iter() {
for bit_idx in (0..8).rev() {
// Reason: 0u8.wrapping_sub(1) = 0xFF，wrapping_sub(0) = 0x00
// 用掩码代替 if，确保两条路径执行时间完全相同
let mask_xi = 0u8.wrapping_sub((byte_xi >> bit_idx) & 1);
for j in 0..16 {
z[j] ^= v[j] & mask_xi;
}
let lsb = v[15] & 1;
for j in (1..16).rev() {
v[j] = (v[j] >> 1) | (v[j - 1] << 7);
}
// Reason: 0u64.wrapping_sub(1) = 0xFFFF...，wrapping_sub(0) = 0x0000...
// 单次 u64 掩码覆盖原来 16 次 u8 掩码操作
let mask = 0u64.wrapping_sub(((byte_xi >> bit_idx) & 1) as u64);
z[0] ^= v[0] & mask;
z[1] ^= v[1] & mask;
// GF(2^128) 右移 1 位（= 乘以 x），带规约多项式 x^128+x^7+x^2+x+1
// Reason: v[0] 的 bit 0（= 大端第 64 位）移入 v[1] 的 bit 63，
// v[1] 的 bit 0（= GF 元素 x^0 系数）移出后触发规约。
let lsb = v[1] & 1;
let carry = v[0] & 1;
v[0] >>= 1;
// Reason: 同上，掩码替代 if lsb == 1，消除 GF 规约的秘密依赖分支
let reduce_mask = 0u8.wrapping_sub(lsb);
v[0] ^= 0xE1 & reduce_mask;
v[1] = (v[1] >> 1) | (carry << 63);
// Reason: 规约项 0xE1_00...00 对应 x^7+x^2+x+1 写入最高字节（v[0] MSB 端），
// 掩码替代 if lsb，执行路径完全相同
let reduce_mask = 0u64.wrapping_sub(lsb);
v[0] ^= 0xE100_0000_0000_0000u64 & reduce_mask;
}
}
z
let mut out = [0u8; 16];
out[0..8].copy_from_slice(&z[0].to_be_bytes());
out[8..16].copy_from_slice(&z[1].to_be_bytes());
out
}
/// GHASH 认证函数（NIST SP 800-38D §6.4）