HackMD2024q1 第 7 週測驗題
目的: 檢驗學員對 bitwise 的認知
作答表單: 測驗 1 (針對 Linux 核心「設計」課程)
作答表單: 測驗 2 (針對 Linux 核心「實作」課程)
測驗 1
SIMD within a register (SWAR) 是軟體最佳化技巧之一，以下展示 SWAR 運用於 64 位元微處理器架構，原本判斷 2 個 32 位元寬度的整數是否都是奇數 (odd)，可能會這樣撰寫:
#include <stdint.h>
bool both_odd(uint32_t x, uint32_t y)
{
return (x & 1) && (y & 1);
}
但我們可先組合 (compound) 2 個 32 位元寬度的整數為 1 個 64 位元整數，再運用特製的 bitmask，從而減少運算量:
static uint64_t SWAR_ODD_MASK = (1L << 32) + 1;
bool both_odd_swar(uint64_t xy) {
return (xy & SWAR_ODD_MASK) == SWAR_ODD_MASK;
}
測試程式:
static inline uint64_t bit_compound(uint32_t x, uint32_t y)
{
return ((0L + x) << 32) | ((y + 0L) & (-1L >> 32));
}
int main()
{
int x = 12345678;
int y = 9012345;
uint64_t xy = bit_compound(x, y);
printf("%d, %d\n", both_odd(x, y), both_odd_swar(xy));
}
延伸閱讀: SIMD and SWAR Techniques
在 Linux 核心原始程式碼中，lib/string.c 具備 memchr 的實作:
/**
* memchr - Find a character in an area of memory.
* @s: The memory area
* @c: The byte to search for
* @n: The size of the area.
*
* returns the address of the first occurrence of @c, or %NULL
* if @c is not found
*/
void *memchr(const void *s, int c, size_t n)
{
const unsigned char *p = s;
while (n-- != 0) {
if ((unsigned char)c == *p++) {
return (void *)(p - 1);
}
}
return NULL;
}
測試程式:
int main()
{
const char str[] = "https://wiki.csie.ncku.edu.tw";
const char ch = '.';
char *ret = memchr_opt(str, ch, strlen(str));
printf("String after |%c| is - |%s|\n", ch, ret);
return 0;
}
預期執行結果:
String after |.| is - |.csie.ncku.edu.tw|
利用上述 SIMD within a register (SWAR) 的技巧，我們可改寫為以下 memchr_opt 函式:
#include <limits.h>
#include <stddef.h>
#include <stdint.h>
#include <string.h>
/* Nonzero if X is not aligned on a "long" boundary */
#define UNALIGNED(X) ((long) X & (sizeof(long) - 1))
/* How many bytes are loaded each iteration of the word copy loop */
#define LBLOCKSIZE (sizeof(long))
/* Threshhold for punting to the bytewise iterator */
#define TOO_SMALL(LEN) ((LEN) < LBLOCKSIZE)
#if LONG_MAX == 2147483647L
#define DETECT_NULL(X) (((X) - (0x01010101)) & ~(X) & (0x80808080))
#else
#if LONG_MAX == 9223372036854775807L
/* Nonzero if X (a long int) contains a NULL byte. */
#define DETECT_NULL(X) \
(((X) - (0x0101010101010101)) & ~(X) & (0x8080808080808080))
#else
#error long int is not a 32bit or 64bit type.
#endif
#endif
/* @return nonzero if (long)X contains the byte used to fill MASK. */
#define DETECT_CHAR(X, mask) DETECT_NULL(AAAA)
void *memchr_opt(const void *str, int c, size_t len)
{
const unsigned char *src = (const unsigned char *) str;
unsigned char d = c;
while (UNALIGNED(src)) {
if (!len--)
return NULL;
if (*src == d)
return (void *) src;
src++;
}
if (!TOO_SMALL(len)) {
/* If we get this far, len is large and src is word-aligned. */
/* The fast code reads the source one word at a time and only performs
* a bytewise search on word-sized segments if they contain the search
* character. This is detected by XORing the word-sized segment with a
* word-sized block of the search character, and then checking for the
* presence of NULL in the result.
*/
unsigned long *asrc = (unsigned long *) src;
unsigned long mask = d << 8 | d;
mask |= mask << 16;
for (unsigned int i = 32; i < LBLOCKSIZE * 8; i <<= 1)
mask |= BBBB;
while (len >= LBLOCKSIZE) {
if (DETECT_CHAR(CCCC, mask))
break;
asrc++;
len -= LBLOCKSIZE;
}
/* If there are fewer than LBLOCKSIZE characters left, then we resort to
* the bytewise loop.
*/
src = (unsigned char *) asrc;
}
while (len--) {
if (*src == d)
return (void *) src;
src++;
}
return NULL;
}
請補完程式碼，使上述 memchr_opt 的實作符合 memchr 行為，作答規範:
- 以最精簡的程式碼撰寫
- 以第一次作業規範的程式碼風格書寫
- AAAA, BBBB, CCCC 皆為表示式
延伸問題:
- 解釋上述程式碼運作原理
- 比較 Linux 核心原本 (與平台無關) 的實作和
memchr_opt，設計實驗來觀察隨著字串長度和特定 pattern 變化的效能影響 - 研讀 2022 年學生報告，在 Linux 核心原始程式碼找出 x86_64 或 arm64 對應的最佳化實作，跟上述程式碼比較，並嘗試舉出其中的策略和分析
下一步：貢獻程式碼到 Linux 核心
Phoronix 報導
測驗 2
若干處理器架構無法有效地處理除法，且我們想避免除法造成的例外狀況 (如 Division by zero)，於是我們考慮以下實作:
#include <stdbool.h>
#include <stdint.h>
#define likely(x) __builtin_expect(!!(x), 1)
static inline bool is_power_of_2(unsigned long n)
{
return (n != 0 && ((n & (n - 1)) == 0));
}
static inline int __ilog2_u32(uint32_t x)
{
return x ? sizeof(x) * 8 - __builtin_clz(x) - 1 : 0;
}
static inline int __ilog2_u64(uint64_t x)
{
uint32_t h = x >> 32;
if (h)
return sizeof(h) * 8 - __builtin_clz(h) + 31;
return x ? sizeof(x) * 8 - __builtin_clz(x) - 1 : 0;
}
/* Calculates the logarithm base 2 of a 32-bit or 64-bit unsigned integer.
* This function performs a constant-capable log base 2 computation, allowing
* it to initialize global variables with constant data, which explains the
* extensive use of ternary operators.
* It automatically chooses the optimized version suitable for the size of 'n',
* based on whether it is 32-bit or 64-bit.
*/
#define ilog2(n) \
(__builtin_constant_p(n) ? ((n) < 2 ? 0 : 63 - __builtin_clzll(n)) \
: (sizeof(n) <= 4) ? __ilog2_u32(n) \
: __ilog2_u64(n))
/* In cases where the divisor is a constant, we compute its inverse during
* compile time. This allows us to replace the division operation with several
* inline multiplications, significantly improving performance.
*/
#define __div64_const32(n, ___b) \
({ \
/* This approach multiplies by the reciprocal of b: to divide n by \
* b, we multiply n by (p / b) and then divide by p. The \
* efficiency comes from the compiler's ability to optimize much \
* of this code during compile time through constant propagation, \
* leaving behind only a few multiplication operations. This is \
* why we use this large macro, as a static inline function \
* does not consistently achieve the same optimization. \
*/ \
uint64_t ___res, ___x, ___t, ___m, ___n = (n); \
bool ___bias; \
\
/* determine MSB of b */ \
uint32_t ___p = 1 << ilog2(___b); \
\
/* compute m = ((p << 64) + b - 1) / b */ \
___m = (~0ULL / ___b) * ___p; \
___m += (((~0ULL % ___b + 1) * ___p) + ___b - 1) / ___b; \
\
/* one less than the dividend with highest result */ \
___x = ~0ULL / ___b * ___b - 1; \
\
/* test our ___m with res = m * x / (p << 64) */ \
___res = ((___m & 0xffffffff) * (___x & 0xffffffff)) >> 32; \
___t = ___res += (___m & 0xffffffff) * (___x >> 32); \
___res += (___x & 0xffffffff) * (___m >> 32); \
___t = (___res < ___t) ? (1ULL << 32) : 0; \
___res = (___res >> 32) + ___t; \
___res += (___m >> 32) * (___x >> 32); \
___res /= ___p; \
\
/* Next, we will clean up and enhance the efficiency of our current \
* implementation. \
*/ \
if (~0ULL % (___b / (___b & -___b)) == 0) { \
/* special case, can be simplified to ... */ \
___n /= (___b & -___b); \
___m = ~0ULL / (___b / (___b & -___b)); \
___p = 1; \
___bias = true; \
} else if (___res != ___x / ___b) { \
/* It is necessary to introduce a bias to counteract errors \
* caused by bit truncation. Without this bias, an extra bit \
* would be required to accurately represent the variable 'm', \
* which would exceed the capacity of a 64-bit variable. To \
* manage this, the solution involves setting 'm' equal to 'p' \
* divided by 'b', and 'n' divided by 'b' as equal to \
* (n * m + m) / p. This approach is taken because avoiding the \
* bias is not feasible due to the limitations imposed by bit \
* truncation errors and the 64-bit variable size constraint. \
*/ \
___bias = true; \
/* Compute m = (p << 64) / b */ \
___m = (~0ULL / ___b) * ___p; \
___m += ((~0ULL % ___b + AAAA) * ___p) / ___b; \
} else { \
/* Reduce m / p, and try to clear bit 31 of m when possible, \
* otherwise that will need extra overflow handling later. \
*/ \
uint32_t ___bits = -(___m & -___m); \
___bits |= ___m >> 32; \
___bits = (~___bits) << 1; \
/* If ___bits == 0 then setting bit 31 is unavoidable. \
* Simply apply the maximum possible reduction in that \
* case. Otherwise the MSB of ___bits indicates the \
* best reduction we should apply. \
*/ \
if (!___bits) { \
___p /= (___m & -___m); \
___m /= (___m & -___m); \
} else { \
BBBB; \
CCCC; \
} \
/* No bias needed. */ \
___bias = false; \
} \
\
/* Now we have a combination of 2 conditions: \
* 1. whether or not we need to apply a bias, and \
* 2. whether or not there might be an overflow in the cross \
* product determined by (___m & ((1 << 63) | (1 << 31))). \
* \
* Select the best way to do (m_bias + m * n) / (1 << 64). \
* From now on there will be actual runtime code generated. \
*/ \
___res = __xprod_64(___m, ___n, ___bias); \
\
___res /= ___p; \
})
/*
* Semantic: retval = ((bias ? m : 0) + m * n) >> 64
*
* The product is a 128-bit value, scaled down to 64 bits.
* Assuming constant propagation to optimize away unused conditional code.
*/
static inline uint64_t __xprod_64(const uint64_t m, uint64_t n, bool bias)
{
uint32_t m_lo = m;
uint32_t m_hi = m >> 32;
uint32_t n_lo = n;
uint32_t n_hi = n >> 32;
uint64_t res;
uint32_t res_lo, res_hi;
if (!bias) {
res = ((uint64_t) m_lo * n_lo) >> 32;
} else if (!(m & ((1ULL << 63) | (1ULL << 31)))) {
/* it is impossible for an overflow to occur in this case */
res = (m + (uint64_t) m_lo * n_lo) >> 32;
} else {
res = m + (uint64_t) m_lo * n_lo;
res_lo = res >> 32;
res_hi = (res_lo < m_hi);
res = res_lo | ((uint64_t) res_hi << 32);
}
if (!(m & ((1ULL << 63) | (1ULL << 31)))) {
/* it is impossible for an overflow to occur in this case */
res += (uint64_t) m_lo * n_hi;
res += (uint64_t) m_hi * n_lo;
res >>= 32;
} else {
res += (uint64_t) m_lo * n_hi;
uint32_t tmp = res >> 32;
res += (uint64_t) m_hi * n_lo;
res_lo = res >> 32;
res_hi = (res_lo < tmp);
res = res_lo | ((uint64_t) res_hi << 32);
}
res += (uint64_t) m_hi * n_hi;
return res;
}
static inline uint32_t __div64_32(uint64_t *n, uint32_t base)
{
uint64_t rem = *n;
uint64_t b = base;
uint64_t res, d = 1;
uint32_t high = rem >> 32;
/* Reduce the thing a bit first */
res = 0;
if (high >= base) {
high /= base;
res = (uint64_t) high << 32;
rem -= (uint64_t) (high * base) << 32;
}
while ((int64_t) b > 0 && b < rem) {
b = b + b;
d = d + d;
}
do {
if (rem >= b) {
rem -= b;
res += d;
}
b >>= 1;
d >>= 1;
} while (d);
*n = res;
return rem;
}
/* The unnecessary pointer compare is there to check for type safety.
* n must be 64bit.
*/
#define div64(n, base) \
({ \
uint32_t __base = (base); \
uint32_t __rem; \
if (__builtin_constant_p(__base) && is_power_of_2(__base)) { \
__rem = (n) & (__base - 1); \
(n) >>= DDDD; \
} else if (__builtin_constant_p(__base) && __base != 0) { \
uint32_t __res_lo, __n_lo = (n); \
(n) = __div64_const32(n, __base); \
/* the remainder can be computed with 32-bit regs */ \
__res_lo = (n); \
__rem = __n_lo - __res_lo * __base; \
} else if (likely(((n) >> 32) == 0)) { \
__rem = (uint32_t) (n) % __base; \
(n) = (uint32_t) (n) / __base; \
} else { \
__rem = __div64_32(&(n), __base); \
} \
__rem; \
})
使用案例:
#include <stdio.h>
int main(int argc, char *argv[])
{
uint64_t res;
res = argc + 5;
div64(res, 3);
printf("%lu\n", res);
}
作答規範:
- 以最精簡的形式書寫，並依循第一次作業的程式碼風格
- AAAA 為整數
- BBBB 為包含
___p的表示式 - CCCC 為包含
___m的表示式 - BBBB, CCC, DDDD 都該包含
ilog2的使用
延伸問題:
- 解釋上述程式碼運作原理
- 在 Linux 核心原始程式碼找到相關的程式碼並設計實驗 (使用 Linux 核心模組) 探討程式碼的效益和限制
