我在C中有一个代码，它将数组的每个元素乘以一个数字(0-9(，得到一系列以10为底的数字。
我的问题是，这个函数的运行时间比我预期的要长。我需要它更快。我知道，在优化函数时，我的问题是进位的依赖性。如何修改此代码以解决此问题并使代码更快？解决方案可以使用内部函数或其他专门技术。
到目前为止，我最快的版本是：
void ConstMult( uint8_t *V, size_t N, uint8_t digit )
{
uint8_t CARRY = 0;
for ( size_t i=0; i< N; ++i )
{
V[i] = V[i] * digit + CARRY;
CARRY = ((uint32_t)V[i] * (uint32_t)0xCCCD) >> 19;
V[i] -= (CARRY << 3) + (CARRY << 1);
}
}
但我也尝试了这些较慢的方法：
uint8_t ConstMult( uint8_t *V, size_t N, uint8_t digit )
{
uint8_t CARRY = 0;
for ( int i=0; i< N; i++ )
{
char R = V[i] * digit + CARRY;
CARRY = R / 10;
R = R - CARRY*10;
V[i] = R;
}
return CARRY; // may be from 0 to 9
}
uint8_t ConstMult(uint8_t *V, size_t N, uint8_t digit)
{
uint8_t CARRY = 0;
uint8_t ja = 0;
for (size_t i = 0; i < N; ++i) {
uint8_t aux = V[i] * digit;
uint8_t R = aux + CARRY;
CARRY = ((u_int32_t)R*(u_int32_t)0xCCCD) >> 19;
ja = (CARRY << 3) + 2*CARRY;
R -= ja;
V[i] = R;
}
return CARRY;
}
这里有另一个实现(比其他实现快得多(：
void ConstMult4(uint8_t *V, size_t N, uint8_t digit)
{
uint8_t CARRY = 0;
const uint32_t coef7 = digit * 10000000;
const uint32_t coef6 = digit * 1000000;
const uint32_t coef5 = digit * 100000;
const uint32_t coef4 = digit * 10000;
const uint32_t coef3 = digit * 1000;
const uint32_t coef2 = digit * 100;
const uint32_t coef1 = digit * 10;
const uint32_t coef0 = digit;
static uint8_t table[10000][4];
static int init = 1;
if(init)
{
for(int i=0 ; i<10000 ; ++i)
{
table[i][0] = (i / 1) % 10;
table[i][1] = (i / 10) % 10;
table[i][2] = (i / 100) % 10;
table[i][3] = (i / 1000) % 10;
}
init = 0;
}
for(size_t i=0 ; i<N/8*8 ; i+=8)
{
const uint32_t val = V[i+7]*coef7 + V[i+6]*coef6 + V[i+5]*coef5 + V[i+4]*coef4 + V[i+3]*coef3 + V[i+2]*coef2 + V[i+1]*coef1 + V[i+0]*coef0 + CARRY;
CARRY = val / 100000000;
const uint32_t loVal = val % 10000;
const uint32_t hiVal = val / 10000 - CARRY * 10000;
const uint8_t* loTablePtr = &table[loVal][0];
const uint8_t* hiTablePtr = &table[hiVal][0];
// Assume the compiler optimize the 2 following calls
// (otherwise the performance could be quite bad).
// memcpy is used to prevent performance issue due to pointer aliasing.
memcpy(V+i, loTablePtr, 4);
memcpy(V+i+4, hiTablePtr, 4);
}
for(size_t i=N/8*8 ; i<N ; ++i)
{
V[i] = V[i] * digit + CARRY;
CARRY = V[i] / 10;
V[i] -= CARRY * 10;
}
}
该实现假设V和digit中的计算数字实际上是数字。它比其他方法快得多：
- 按照@phuclv的建议，在内部使用更大的基数(它减少了关键路径并引入了更多的并行性(
- 使用@chqrlieforyellowblockquotes提出的查找表(它可以非常快速地计算除法/模运算(
使用SSE 4.1内部函数(SIMD指令(甚至可以改进此代码。但代价是不太可移植的代码(尽管它可以在大多数基于x86_64的现代处理器上工作(。以下是实现：
void ConstMult5(uint8_t *V, size_t N, uint8_t digit)
{
uint8_t CARRY = 0;
static uint8_t table[10000][4];
static int init = 1;
if(init)
{
for(int i=0 ; i<10000 ; ++i)
{
table[i][0] = (i / 1) % 10;
table[i][1] = (i / 10) % 10;
table[i][2] = (i / 100) % 10;
table[i][3] = (i / 1000) % 10;
}
init = 0;
}
__m128i coefs1 = _mm_set_epi16(1000, 100, 10, 1, 1000, 100, 10, 1);
__m128i coefs2 = _mm_set_epi32(10000*digit, 10000*digit, digit, digit);
for(size_t i=0 ; i<N/16*16 ; i+=8)
{
// Require SSE 4.1 (thus smmintrin.h need to be included)
const __m128i vBlock = _mm_loadu_si128((const __m128i*)&V[i]); // load 16 x uint8_t values (only half is used)
const __m128i v = _mm_cvtepu8_epi16(vBlock); // Convert the block to 8 x int16_t values
const __m128i tmp1 = _mm_madd_epi16(v, coefs1); // Compute the sum of adjacent pairs of v * coefs1 and put this in 4 x int32_t values
const __m128i tmp2 = _mm_add_epi32(tmp1, _mm_shuffle_epi32(tmp1, 0b10110001)); // Horizontal partial sum of 4 x int32_t values
const __m128i tmp3 = _mm_mul_epu32(tmp2, coefs2); // Compute tmp2 * coefs2 and put this in 2 x int64_t values
const uint32_t val = _mm_extract_epi64(tmp3, 1) + _mm_extract_epi64(tmp3, 0) + CARRY; // Final horizontal sum with CARRY
CARRY = val / 100000000;
const uint32_t loVal = val % 10000;
const uint32_t hiVal = val / 10000 - CARRY * 10000;
const uint8_t* loTablePtr = &table[loVal][0];
const uint8_t* hiTablePtr = &table[hiVal][0];
// See the memcpy remark in the code above (alternative version).
memcpy(V+i, loTablePtr, 4);
memcpy(V+i+4, hiTablePtr, 4);
}
for(size_t i=N/16*16 ; i<N ; ++i)
{
V[i] = V[i] * digit + CARRY;
CARRY = V[i] / 10;
V[i] -= CARRY * 10;
}
}
以下是我的机器(使用i7-9700KF处理器(的性能结果(使用随机输入在1000次运行中重复并平均(：
ConstMult0(10000): 11.702 us
ConstMult3(10000): 6.768 us (last optimized version)
ConstMult4(10000): 3.569 us
ConstMult5(10000): 2.552 us
最终的基于SSE的版本比原来的实现快4.6倍！
这里有一个函数，它使用辅助表一次处理块2个字节，而不进行分割：
uint8_t ConstMult3(uint8_t *V, size_t N, uint8_t digit) {
#define TABLE_SIZE ((9 * 256 + 9) * 9 + 9 + 1)
static uint32_t table[TABLE_SIZE];
if (!table[1]) {
for (uint32_t x = 0; x < TABLE_SIZE; x++) {
uint32_t u = x % 256 % 10;
uint32_t d = (x / 256 + x % 256 / 10) % 10;
uint32_t c = (x / 256 + x % 256 / 10) / 10;
//table[x] = u | (d << 8) | (c << 16);
// modified following Jerome Richard's comment
table[x] = c | (u << 8) | (d << 16);
}
}
if (N == 0 || digit <= 1) {
if (digit == 0)
memset(V, 0, N);
return 0;
} else {
size_t CARRY = 0;
if ((uintptr_t)V & 1) { // V is misaligned
int R = V[0] * digit + (uint8_t)CARRY;
CARRY = (uint8_t)(R / 10);
V[0] = (uint8_t)(R - CARRY * 10);
V++;
N--;
}
{ // handle aligned block 2 bytes at a time
uint16_t *V2 = (uint16_t *)(void *)V;
size_t N2 = N / 2;
for (size_t i = 0; i < N2; i++) {
uint32_t x = table[V2[i] * digit + CARRY];
//V2[i] = (uint16_t)x;
//CARRY = x >> 16;
// modified following Jerome Richard's comment
V2[i] = (uint16_t)(x >> 8);
CARRY = (uint8_t)x;
}
}
if (N & 1) { // handle last byte
int R = V[N - 1] * digit + (uint8_t)CARRY;
CARRY = (uint8_t)(R / 10);
V[N - 1] = (uint8_t)(R - CARRY * 10);
}
return (uint8_t)CARRY;
}
#undef TABLE_SIZE
}
在我的慢速笔记本电脑上，在64位模式下使用clang 9.0，我得到了ConstMult0、ConstMult1和ConstMult2的这些计时
ConstMult0(1000000(：15.816ms sum0=4495507，sum=4501418ConstMult1(1000000(：16.4464ms sum0=4495507，sum=4501418ConstMult2(1000000(：16.483ms sum0=4495507，sum=4501418ConstMult3(1000000(：9.644ms sum0=4495507，sum=4501418
编辑：根据Jérôme Richard的评论，表格内容的一个小变化将额外提高11%的性能：
ConstMult0(1000000(：15.837ms sum0=4500384，sum=44495487ConstMult1(1000000(：16.494ms sum0=4500384，sum=4495487ConstMult2(1000000(：16.482毫秒sum0=4500384，sum=44495487ConstMult3(1000000(：8.537ms sum0=4500384，sum=44495487