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注意:我希望能得到更多有关如何处理和提出此类解决方案的指南,而不是解决方案本身。
我的系统中有一个非常关键的功能,它在特定上下文中显示为排名第一的分析热点。它处于 k-means 迭代的中间(已经是多线程的,使用并行处理每个工作线程中点的子范围)。
ClusterPoint& pt = points[j];
pt.min_index = -1;
pt.min_dist = numeric_limits<float>::max();
for (int i=0; i < num_centroids; ++i)
{
const ClusterCentroid& cent = centroids[i];
const float dist = ...;
if (dist < pt.min_dist) // <-- #1 hotspot
{
pt.min_dist = dist;
pt.min_index = i;
}
}
处理这部分代码所需的任何时间节省都非常重要,所以我经常摆弄它。例如,可能值得将质心循环放在外面,并为给定的质心并行迭代这些点。这里的聚类点数以百万计,而质心数以千计。该算法适用于少数迭代(通常在 10 次以下)。它不寻求完美的收敛/稳定性,只是一些“合理”的近似。
000007FEEE3FB8A1 jl thread_partition+70h (7FEEE3FB780h)
{
ClusterPoint& pt = points[j];
pt.min_index = -1;
pt.min_dist = numeric_limits<float>::max();
for (int i = 0; i < num_centroids; ++i)
000007FEEE3FB8A7 cmp ecx,r10d
000007FEEE3FB8AA jge thread_partition+1F4h (7FEEE3FB904h)
000007FEEE3FB8AC lea rax,[rbx+rbx*2]
000007FEEE3FB8B0 add rax,rax
000007FEEE3FB8B3 lea r8,[rbp+rax*8+8]
{
const ClusterCentroid& cent = centroids[i];
const float x = pt.pos[0] - cent.pos[0];
const float y = pt.pos[1] - cent.pos[1];
000007FEEE3FB8B8 movss xmm0,dword ptr [rdx]
const float z = pt.pos[2] - cent.pos[2];
000007FEEE3FB8BC movss xmm2,dword ptr [rdx+4]
000007FEEE3FB8C1 movss xmm1,dword ptr [rdx-4]
000007FEEE3FB8C6 subss xmm2,dword ptr [r8]
000007FEEE3FB8CB subss xmm0,dword ptr [r8-4]
000007FEEE3FB8D1 subss xmm1,dword ptr [r8-8]
const float dist = x*x + y*y + z*z;
000007FEEE3FB8D7 mulss xmm2,xmm2
000007FEEE3FB8DB mulss xmm0,xmm0
000007FEEE3FB8DF mulss xmm1,xmm1
000007FEEE3FB8E3 addss xmm2,xmm0
000007FEEE3FB8E7 addss xmm2,xmm1
if (dist < pt.min_dist)
// VTUNE HOTSPOT
000007FEEE3FB8EB comiss xmm2,dword ptr [rdx-8]
000007FEEE3FB8EF jae thread_partition+1E9h (7FEEE3FB8F9h)
{
pt.min_dist = dist;
000007FEEE3FB8F1 movss dword ptr [rdx-8],xmm2
pt.min_index = i;
000007FEEE3FB8F6 mov dword ptr [rdx-10h],ecx
000007FEEE3FB8F9 inc ecx
000007FEEE3FB8FB add r8,30h
000007FEEE3FB8FF cmp ecx,r10d
000007FEEE3FB902 jl thread_partition+1A8h (7FEEE3FB8B8h)
for (int j = *irange.first; j < *irange.last; ++j)
000007FEEE3FB904 inc edi
000007FEEE3FB906 add rdx,20h
000007FEEE3FB90A cmp edi,dword ptr [rsi+4]
000007FEEE3FB90D jl thread_partition+31h (7FEEE3FB741h)
000007FEEE3FB913 mov rbx,qword ptr [irange]
}
}
}
}
我们被迫瞄准 SSE 2 —— 有点落后于我们的时代,但是当我们假设即使 SSE 4 也可以作为最低要求时(用户有一些原型(prototype) Intel 机器),用户群实际上被绊倒了一次。
#define _SECURE_SCL 0
#include <iostream>
#include <fstream>
#include <vector>
#include <limits>
#include <ctime>
#if defined(_MSC_VER)
#define ALIGN16 __declspec(align(16))
#else
#include <malloc.h>
#define ALIGN16 __attribute__((aligned(16)))
#endif
using namespace std;
// Aligned memory allocation (for SIMD).
static void* malloc16(size_t amount)
{
#ifdef _MSC_VER
return _aligned_malloc(amount, 16);
#else
void* mem = 0;
posix_memalign(&mem, 16, amount);
return mem;
#endif
}
template <class T>
static T* malloc16_t(size_t num_elements)
{
return static_cast<T*>(malloc16(num_elements * sizeof(T)));
}
// Aligned free.
static void free16(void* mem)
{
#ifdef _MSC_VER
return _aligned_free(mem);
#else
free(mem);
#endif
}
// Test parameters.
enum {num_centroids = 512};
enum {num_points = num_centroids * 2000};
enum {num_iterations = 5};
static const float range = 10.0f;
class Points
{
public:
Points(): data(malloc16_t<Point>(num_points))
{
for (int p=0; p < num_points; ++p)
{
const float xyz[3] =
{
range * static_cast<float>(rand()) / RAND_MAX,
range * static_cast<float>(rand()) / RAND_MAX,
range * static_cast<float>(rand()) / RAND_MAX
};
init(p, xyz);
}
}
~Points()
{
free16(data);
}
void init(int n, const float* xyz)
{
data[n].centroid = -1;
data[n].xyz[0] = xyz[0];
data[n].xyz[1] = xyz[1];
data[n].xyz[2] = xyz[2];
}
void associate(int n, int new_centroid)
{
data[n].centroid = new_centroid;
}
int centroid(int n) const
{
return data[n].centroid;
}
float* operator[](int n)
{
return data[n].xyz;
}
private:
Points(const Points&);
Points& operator=(const Points&);
struct Point
{
int centroid;
float xyz[3];
};
Point* data;
};
class Centroids
{
public:
Centroids(Points& points): data(malloc16_t<Centroid>(num_centroids))
{
// Naive initial selection algorithm, but outside the
// current area of interest.
for (int c=0; c < num_centroids; ++c)
init(c, points[c]);
}
~Centroids()
{
free16(data);
}
void init(int n, const float* xyz)
{
data[n].count = 0;
data[n].xyz[0] = xyz[0];
data[n].xyz[1] = xyz[1];
data[n].xyz[2] = xyz[2];
}
void reset(int n)
{
data[n].count = 0;
data[n].xyz[0] = 0.0f;
data[n].xyz[1] = 0.0f;
data[n].xyz[2] = 0.0f;
}
void sum(int n, const float* pt_xyz)
{
data[n].xyz[0] += pt_xyz[0];
data[n].xyz[1] += pt_xyz[1];
data[n].xyz[2] += pt_xyz[2];
++data[n].count;
}
void average(int n)
{
if (data[n].count > 0)
{
const float inv_count = 1.0f / data[n].count;
data[n].xyz[0] *= inv_count;
data[n].xyz[1] *= inv_count;
data[n].xyz[2] *= inv_count;
}
}
float* operator[](int n)
{
return data[n].xyz;
}
int find_nearest(const float* pt_xyz) const
{
float min_dist_squared = numeric_limits<float>::max();
int min_centroid = -1;
for (int c=0; c < num_centroids; ++c)
{
const float* cen_xyz = data[c].xyz;
const float x = pt_xyz[0] - cen_xyz[0];
const float y = pt_xyz[1] - cen_xyz[1];
const float z = pt_xyz[2] - cen_xyz[2];
const float dist_squared = x*x + y*y * z*z;
if (min_dist_squared > dist_squared)
{
min_dist_squared = dist_squared;
min_centroid = c;
}
}
return min_centroid;
}
private:
Centroids(const Centroids&);
Centroids& operator=(const Centroids&);
struct Centroid
{
int count;
float xyz[3];
};
Centroid* data;
};
// A high-precision real timer would be nice, but we lack C++11 and
// the coarseness of the testing here should allow this to suffice.
static double sys_time()
{
return static_cast<double>(clock()) / CLOCKS_PER_SEC;
}
static void k_means(Points& points, Centroids& centroids)
{
// Find the closest centroid for each point.
for (int p=0; p < num_points; ++p)
{
const float* pt_xyz = points[p];
points.associate(p, centroids.find_nearest(pt_xyz));
}
// Reset the data of each centroid.
for (int c=0; c < num_centroids; ++c)
centroids.reset(c);
// Compute new position sum of each centroid.
for (int p=0; p < num_points; ++p)
centroids.sum(points.centroid(p), points[p]);
// Compute average position of each centroid.
for (int c=0; c < num_centroids; ++c)
centroids.average(c);
}
int main()
{
Points points;
Centroids centroids(points);
cout << "Starting simulation..." << endl;
double start_time = sys_time();
for (int i=0; i < num_iterations; ++i)
k_means(points, centroids);
cout << "Time passed: " << (sys_time() - start_time) << " secs" << endl;
cout << "# Points: " << num_points << endl;
cout << "# Centroids: " << num_centroids << endl;
// Write the centroids to a file to give us some crude verification
// of consistency as we make changes.
ofstream out("centroids.txt");
for (int c=0; c < num_centroids; ++c)
out << "Centroid " << c << ": " << centroids[c][0] << "," << centroids[c][1] << "," << centroids[c][2] << endl;
}
我知道表面测试的危险性,但由于它已经被认为是以前真实世界 session 的热点,我希望这是可以原谅的。我也只对与微优化此类代码相关的一般技术感兴趣。
min_dist
成员并使其成为局部变量)。质心与点之间的确切比率也有点不同,但希望足够接近以将此处的改进转化为原始代码。在这个肤浅的测试中它也是单线程的,反汇编看起来很不一样,所以我可能会冒险在没有原始版本的情况下优化这个肤浅的测试(我现在愿意承担这个风险,因为我对扩展我的知识更感兴趣可以优化这些情况的技术,而不是针对这种情况的解决方案)。
-if (min_dist_squared > dist_squared)
-{
- min_dist_squared = dist_squared;
- pt.centroid = c;
-}
有了这个:
+const bool found_closer = min_dist_squared > dist_squared;
+pt.centroid = bitselect(found_closer, c, pt.centroid);
+min_dist_squared = bitselect(found_closer, dist_squared, min_dist_squared);
.. 只是发现时间从 ~5.6 秒上升到 ~12.5 秒。然而,这不是他的错,也不会影响他的解决方案的值(value)——这是我未能理解机器级别真正发生的事情并在黑暗中刺伤的原因。那个显然错过了,显然我不是我最初认为的分支预测错误的受害者。尽管如此,他提出的解决方案是在这种情况下尝试的绝妙且通用的功能,我很高兴将其添加到我的提示和技巧工具箱中。现在进行第 2 轮。
// New version of Centroids::find_nearest (from harold's solution):
int find_nearest(const float* pt_xyz) const
{
__m128i min_index = _mm_set_epi32(3, 2, 1, 0);
__m128 xdif = _mm_sub_ps(_mm_set1_ps(pt_xyz[0]), _mm_load_ps(cen_x));
__m128 ydif = _mm_sub_ps(_mm_set1_ps(pt_xyz[1]), _mm_load_ps(cen_y));
__m128 zdif = _mm_sub_ps(_mm_set1_ps(pt_xyz[2]), _mm_load_ps(cen_z));
__m128 min_dist = _mm_add_ps(_mm_add_ps(_mm_mul_ps(xdif, xdif),
_mm_mul_ps(ydif, ydif)),
_mm_mul_ps(zdif, zdif));
__m128i index = min_index;
for (int i=4; i < num_centroids; i += 4)
{
xdif = _mm_sub_ps(_mm_set1_ps(pt_xyz[0]), _mm_load_ps(cen_x + i));
ydif = _mm_sub_ps(_mm_set1_ps(pt_xyz[1]), _mm_load_ps(cen_y + i));
zdif = _mm_sub_ps(_mm_set1_ps(pt_xyz[2]), _mm_load_ps(cen_z + i));
__m128 dist = _mm_add_ps(_mm_add_ps(_mm_mul_ps(xdif, xdif),
_mm_mul_ps(ydif, ydif)),
_mm_mul_ps(zdif, zdif));
__m128i mask = _mm_castps_si128(_mm_cmplt_ps(dist, min_dist));
min_dist = _mm_min_ps(min_dist, dist);
min_index = _mm_or_si128(_mm_and_si128(index, mask),
_mm_andnot_si128(mask, min_index));
index = _mm_add_epi32(index, _mm_set1_epi32(4));
}
ALIGN16 float mdist[4];
ALIGN16 uint32_t mindex[4];
_mm_store_ps(mdist, min_dist);
_mm_store_si128((__m128i*)mindex, min_index);
float closest = mdist[0];
int closest_i = mindex[0];
for (int i=1; i < 4; i++)
{
if (mdist[i] < closest)
{
closest = mdist[i];
closest_i = mindex[i];
}
}
return closest_i;
}
Harold 的 SIMD 解决方案(更正) - ~2.5 秒
bitselect
技巧的所有贡献!我希望我能接受所有的答案。我可能最终会在某个时候尝试所有这些,但现在我的功课是理解这些非算术 SIMD 操作中的一些。
int find_nearest_simd(const float* pt_xyz) const
{
__m128i min_index = _mm_set_epi32(3, 2, 1, 0);
__m128 pt_xxxx = _mm_set1_ps(pt_xyz[0]);
__m128 pt_yyyy = _mm_set1_ps(pt_xyz[1]);
__m128 pt_zzzz = _mm_set1_ps(pt_xyz[2]);
__m128 xdif = _mm_sub_ps(pt_xxxx, _mm_load_ps(cen_x));
__m128 ydif = _mm_sub_ps(pt_yyyy, _mm_load_ps(cen_y));
__m128 zdif = _mm_sub_ps(pt_zzzz, _mm_load_ps(cen_z));
__m128 min_dist = _mm_add_ps(_mm_add_ps(_mm_mul_ps(xdif, xdif),
_mm_mul_ps(ydif, ydif)),
_mm_mul_ps(zdif, zdif));
__m128i index = min_index;
for (int i=4; i < num_centroids; i += 4)
{
xdif = _mm_sub_ps(pt_xxxx, _mm_load_ps(cen_x + i));
ydif = _mm_sub_ps(pt_yyyy, _mm_load_ps(cen_y + i));
zdif = _mm_sub_ps(pt_zzzz, _mm_load_ps(cen_z + i));
__m128 dist = _mm_add_ps(_mm_add_ps(_mm_mul_ps(xdif, xdif),
_mm_mul_ps(ydif, ydif)),
_mm_mul_ps(zdif, zdif));
index = _mm_add_epi32(index, _mm_set1_epi32(4));
__m128i mask = _mm_castps_si128(_mm_cmplt_ps(dist, min_dist));
min_dist = _mm_min_ps(min_dist, dist);
min_index = _mm_or_si128(_mm_and_si128(index, mask),
_mm_andnot_si128(mask, min_index));
}
ALIGN16 float mdist[4];
ALIGN16 uint32_t mindex[4];
_mm_store_ps(mdist, min_dist);
_mm_store_si128((__m128i*)mindex, min_index);
float closest = mdist[0];
int closest_i = mindex[0];
for (int i=1; i < 4; i++)
{
if (mdist[i] < closest)
{
closest = mdist[i];
closest_i = mindex[i];
}
}
return closest_i;
}
最佳答案
太糟糕了,我们不能使用 SSE4.1,但很好,SSE2 是。我还没有测试过这个,只是编译它以查看是否存在语法错误并查看程序集是否有意义(这基本上没问题,尽管 GCC 溢出 min_index
即使有些 xmm
寄存器未使用,不知道为什么会发生这种情况)
int find_closest(float *x, float *y, float *z,
float pt_x, float pt_y, float pt_z, int n) {
__m128i min_index = _mm_set_epi32(3, 2, 1, 0);
__m128 xdif = _mm_sub_ps(_mm_set1_ps(pt_x), _mm_load_ps(x));
__m128 ydif = _mm_sub_ps(_mm_set1_ps(pt_y), _mm_load_ps(y));
__m128 zdif = _mm_sub_ps(_mm_set1_ps(pt_z), _mm_load_ps(z));
__m128 min_dist = _mm_add_ps(_mm_add_ps(_mm_mul_ps(xdif, xdif),
_mm_mul_ps(ydif, ydif)),
_mm_mul_ps(zdif, zdif));
__m128i index = min_index;
for (int i = 4; i < n; i += 4) {
xdif = _mm_sub_ps(_mm_set1_ps(pt_x), _mm_load_ps(x + i));
ydif = _mm_sub_ps(_mm_set1_ps(pt_y), _mm_load_ps(y + i));
zdif = _mm_sub_ps(_mm_set1_ps(pt_z), _mm_load_ps(z + i));
__m128 dist = _mm_add_ps(_mm_add_ps(_mm_mul_ps(xdif, xdif),
_mm_mul_ps(ydif, ydif)),
_mm_mul_ps(zdif, zdif));
index = _mm_add_epi32(index, _mm_set1_epi32(4));
__m128i mask = _mm_castps_si128(_mm_cmplt_ps(dist, min_dist));
min_dist = _mm_min_ps(min_dist, dist);
min_index = _mm_or_si128(_mm_and_si128(index, mask),
_mm_andnot_si128(mask, min_index));
}
float mdist[4];
_mm_store_ps(mdist, min_dist);
uint32_t mindex[4];
_mm_store_si128((__m128i*)mindex, min_index);
float closest = mdist[0];
int closest_i = mindex[0];
for (int i = 1; i < 4; i++) {
if (mdist[i] < closest) {
closest = mdist[i];
closest_i = mindex[i];
}
}
return closest_i;
}
min_index = _mm_or_si128(_mm_and_si128(index, mask),
_mm_andnot_si128(mask, min_index));
min_index = _mm_blendv_epi8(min_index, index, mask);
bits 64
section .data
align 16
centroid_four:
dd 4, 4, 4, 4
centroid_index:
dd 0, 1, 2, 3
section .text
global find_closest
proc_frame find_closest
;
; arguments:
; ecx: number of points (multiple of 4 and at least 4)
; rdx -> array of 3 pointers to floats (x, y, z) (the points)
; r8 -> array of 3 floats (the reference point)
;
alloc_stack 0x58
save_xmm128 xmm6, 0
save_xmm128 xmm7, 16
save_xmm128 xmm8, 32
save_xmm128 xmm9, 48
[endprolog]
movss xmm0, [r8]
shufps xmm0, xmm0, 0
movss xmm1, [r8 + 4]
shufps xmm1, xmm1, 0
movss xmm2, [r8 + 8]
shufps xmm2, xmm2, 0
; pointers to x, y, z in r8, r9, r10
mov r8, [rdx]
mov r9, [rdx + 8]
mov r10, [rdx + 16]
; reference point is in xmm0, xmm1, xmm2 (x, y, z)
movdqa xmm3, [rel centroid_index] ; min_index
movdqa xmm4, xmm3 ; current index
movdqa xmm9, [rel centroid_four] ; index increment
paddd xmm4, xmm9
; calculate initial min_dist, xmm5
movaps xmm5, [r8]
subps xmm5, xmm0
movaps xmm7, [r9]
subps xmm7, xmm1
movaps xmm8, [r10]
subps xmm8, xmm2
mulps xmm5, xmm5
mulps xmm7, xmm7
mulps xmm8, xmm8
addps xmm5, xmm7
addps xmm5, xmm8
add r8, 16
add r9, 16
add r10, 16
sub ecx, 4
jna _tail
_loop:
movaps xmm6, [r8]
subps xmm6, xmm0
movaps xmm7, [r9]
subps xmm7, xmm1
movaps xmm8, [r10]
subps xmm8, xmm2
mulps xmm6, xmm6
mulps xmm7, xmm7
mulps xmm8, xmm8
addps xmm6, xmm7
addps xmm6, xmm8
add r8, 16
add r9, 16
add r10, 16
movaps xmm7, xmm6
cmpps xmm6, xmm5, 1
minps xmm5, xmm7
movdqa xmm7, xmm6
pand xmm6, xmm4
pandn xmm7, xmm3
por xmm6, xmm7
movdqa xmm3, xmm6
paddd xmm4, xmm9
sub ecx, 4
ja _loop
_tail:
; calculate horizontal minumum
pshufd xmm0, xmm5, 0xB1
minps xmm0, xmm5
pshufd xmm1, xmm0, 0x4E
minps xmm0, xmm1
; find index of the minimum
cmpps xmm0, xmm5, 0
movmskps eax, xmm0
bsf eax, eax
; index into xmm3, sort of
movaps [rsp + 64], xmm3
mov eax, [rsp + 64 + rax * 4]
movaps xmm9, [rsp + 48]
movaps xmm8, [rsp + 32]
movaps xmm7, [rsp + 16]
movaps xmm6, [rsp]
add rsp, 0x58
ret
endproc_frame
extern "C" int find_closest(int n, float** points, float* reference_point);
关于c++ - 无分支 K 均值(或其他优化),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30023245/
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您好,我正在使用 Matlab 优化求解器,但程序有问题。我收到此消息 fmincon 已停止,因为目标函数值小于目标函数限制的默认值,并且约束满足在约束容差的默认值范围内。我也收到以下消息。警告:矩
处理Visual Studio optimizations的问题为我节省了大量启动和使用它的时间 当我必须进行 J2EE 开发时,我很难回到 Eclipse。因此,我还想知道人们是否有任何提示或技巧可
情况如下:在我的 Excel 工作表中,有一列包含 1-name 形式的条目。考虑到数字也可以是两位数,我想删除这些数字。这本身不是问题,我让它工作了,只是性能太糟糕了。现在我的程序每个单元格输入大约
这样做有什么区别吗: $(".topHorzNavLink").click(function() { var theHoverContainer = $("#hoverContainer");
这个问题已经有答案了: 已关闭11 年前。 Possible Duplicate: What is the cost of '$(this)'? 我经常在一些开发人员代码中看到$(this)引用同一个
我刚刚结束了一个大型开发项目。我们的时间紧迫,因此很多优化被“推迟”。既然我们已经达到了最后期限,我们将回去尝试优化事情。 我的问题是:优化 jQuery 网站时您要寻找的最重要的东西是什么。或者,我
所以我一直在用 JavaScript 编写游戏(不是网络游戏,而是使用 JavaScript 恰好是脚本语言的游戏引擎)。不幸的是,游戏引擎的 JavaScript 引擎是 SpiderMonkey
这是我在正在构建的页面中使用的 SQL 查询。它目前运行大约 8 秒并返回 12000 条记录,这是正确的,但我想知道您是否可以就如何使其更快提出可能的建议? SELECT DISTINCT Adve
如何优化这个? SELECT e.attr_id, e.sku, a.value FROM product_attr AS e, product_attr_text AS a WHERE e.attr
我正在使用这样的结构来测试是否按下了所需的键: def eventFilter(self, tableView, event): if event.type() == QtCore.QEven
我正在使用 JavaScript 从给定的球员列表中计算出羽毛球 double 比赛的所有组合。每个玩家都与其他人组队。 EG。如果我有以下球员a、b、c、d。它们的组合可以是: a & b V c
我似乎无法弄清楚如何让这个 JS 工作。 scroll function 起作用但不能隐藏。还有没有办法用更少的代码行来做到这一点?我希望 .down-arrow 在 50px 之后 fade out
我的问题是关于用于生产的高级优化级联样式表 (CSS) 文件。 多么最新和最完整(准备在实时元素中使用)的 css 优化器/最小化器,它们不仅提供删除空格和换行符,还提供高级功能,如删除过多的属性、合
我读过这个: 浏览器检索在 中请求的所有资源开始呈现 之前的 HTML 部分.如果您将请求放在 中section 而不是,那么页面呈现和下载资源可以并行发生。您应该从 移动尽可能多的资源请求。
我正在处理一些现有的 C++ 代码,这些代码看起来写得不好,而且调用频率很高。我想知道我是否应该花时间更改它,或者编译器是否已经在优化问题。 我正在使用 Visual Studio 2008。 这是一
我正在尝试使用 OpenGL 渲染 3 个四边形(1 个背景图,2 个 Sprite )。我有以下代码: void GLRenderer::onDrawObjects(long p_dt) {
我确实有以下声明: isEnabled = false; if(foo(arg) && isEnabled) { .... } public boolean foo(arg) { some re
(一)深入浅出理解索引结构 实际上,您可以把索引理解为一种特殊的目录。微软的SQL SERVER提供了两种索引:聚集索引(clustered index,也称聚类索引、簇集索引)和非聚集索引(no
一、写在前面 css的优化方案,之前没有提及,所以接下来进行总结一下。 二、具体优化方案 2.1、加载性能 1、css压缩:将写好的css进行打包,可以减少很多的体积。 2、css单一样式:在需要下边
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