`std :: list<> :: sort()`-为什么突然转向自上而下的策略? [英] `std::list<>::sort()` - why the sudden switch to top-down strategy?
问题描述
我记得，自开始以来，最流行的实现 至少这就是GCC C ++标准库实现中的方式(例如，参见解决方案std::list<>::sort()的方法是自下而上的方式(另请参见
请注意，此答案已更新，可以解决以下注释中以及问题之后提到的所有问题，方法是将一系列列表更改为迭代器数组，同时保留了更快的自下而上的合并排序算法，并消除了由于使用自上而下的合并排序算法进行递归而导致堆栈溢出的可能性很小.
我最初不考虑迭代器的原因是由于VS2015自上而下的更改，使我相信尝试更改现有的自下而上算法以使用迭代器存在一些问题.只有当我尝试自己分析对迭代器的转换时，我才意识到有解决方案.
在@sbi的评论中，他询问自上而下的方法的作者Stephan T. Lavavej为何进行了更改. Stephan的回应是避免内存分配和默认构造分配器". VS2015引入了不可默认构造的有状态分配器，这在使用先前版本的列表数组时会出现问题，因为列表的每个实例都会分配一个虚拟节点，并且需要进行更改以处理默认分配器.
Lavavej的解决方案是切换到使用迭代器来跟踪原始列表内而不是内部列表内的运行边界.合并逻辑已更改为使用3个迭代器参数，第一个参数是迭代器，以左行开始，第二个参数是迭代器，以左行结束==迭代器，以右行开始，第三个参数是迭代器，以右行结束.合并过程在合并操作期间使用std :: list :: splice在原始列表内移动节点.这具有异常安全的附加好处.如果调用方的compare函数引发异常，则将对列表进行重新排序，但不会发生数据丢失(假定拼接不会失败).使用先前的方案，如果发生异常，某些(或大部分)数据将在列表的内部数组中，并且数据将从原始列表中丢失.
但是，不需要切换到自上而下的合并排序.最初，我认为VS2015切换到自上而下的原因是未知的，我专注于以与std :: list :: splice相同的方式使用内部接口.后来，我决定研究自下而上的开关，以使用迭代器数组.我意识到存储在内部数组中的运行顺序是最新的(array [0] =最右边)到最旧的(array [last] =最左边)，并且它可以使用与VS2015的自上而下方法相同的基于迭代器的合并逻辑.
对于自下而上的合并排序，array [i]是具有2 ^ i个节点的已排序子列表开头的迭代器，或者为空(使用std :: list :: end表示为空).每个排序后的子列表的末尾将是数组中下一个先前的非空条目中排序后的子列表的开始，或者如果是数组的开始，则是本地迭代器中的排序后的子列表的开始(它指向最新的末尾)跑).与自顶向下方法类似，迭代器数组仅用于跟踪原始链表中已排序的运行边界，而合并过程使用std :: list :: splice在原始链表中移动节点.
如果链表很大并且节点分散，那么将有很多缓存未命中.自下而上的速度将比自上而下的速度快约30％(相当于说自上而下的速度比自下而上的速度慢约42％).再说一次，如果有足够的内存，通常将列表移至数组或向量，对数组或向量进行排序，然后从排序后的数组或向量创建新列表通常会更快.
示例C ++代码:
#define ASZ 32
template <typename T>
void SortList(std::list<T> &ll)
{
if (ll.size() < 2) // return if nothing to do
return;
std::list<T>::iterator ai[ASZ]; // array of iterators
std::list<T>::iterator mi; // middle iterator (end lft, bgn rgt)
std::list<T>::iterator ei; // end iterator
size_t i;
for (i = 0; i < ASZ; i++) // "clear" array
ai[i] = ll.end();
// merge nodes into array
for (ei = ll.begin(); ei != ll.end();) {
mi = ei++;
for (i = 0; (i < ASZ) && ai[i] != ll.end(); i++) {
mi = Merge(ll, ai[i], mi, ei);
ai[i] = ll.end();
}
if(i == ASZ)
i--;
ai[i] = mi;
}
// merge array into single list
ei = ll.end();
for(i = 0; (i < ASZ) && ai[i] == ei; i++);
mi = ai[i++];
while(1){
for( ; (i < ASZ) && ai[i] == ei; i++);
if (i == ASZ)
break;
mi = Merge(ll, ai[i++], mi, ei);
}
}
template <typename T>
typename std::list<T>::iterator Merge(std::list<T> &ll,
typename std::list<T>::iterator li,
typename std::list<T>::iterator mi,
typename std::list<T>::iterator ei)
{
std::list<T>::iterator ni;
(*mi < *li) ? ni = mi : ni = li;
while(1){
if(*mi < *li){
ll.splice(li, ll, mi++);
if(mi == ei)
return ni;
} else {
if(++li == mi)
return ni;
}
}
}
VS2019的std :: list :: sort()的示例替换代码(合并逻辑被制成了单独的内部函数，因为现在已在两个地方使用它了.)
private:
template <class _Pr2>
iterator _Merge(_Pr2 _Pred, iterator _First, iterator _Mid, iterator _Last){
iterator _Newfirst = _First;
for (bool _Initial_loop = true;;
_Initial_loop = false) { // [_First, _Mid) and [_Mid, _Last) are sorted and non-empty
if (_DEBUG_LT_PRED(_Pred, *_Mid, *_First)) { // consume _Mid
if (_Initial_loop) {
_Newfirst = _Mid; // update return value
}
splice(_First, *this, _Mid++);
if (_Mid == _Last) {
return _Newfirst; // exhausted [_Mid, _Last); done
}
}
else { // consume _First
++_First;
if (_First == _Mid) {
return _Newfirst; // exhausted [_First, _Mid); done
}
}
}
}
template <class _Pr2>
void _Sort(iterator _First, iterator _Last, _Pr2 _Pred,
size_type _Size) { // order [_First, _Last), using _Pred, return new first
// _Size must be distance from _First to _Last
if (_Size < 2) {
return; // nothing to do
}
const size_t _ASZ = 32; // array size
iterator _Ai[_ASZ]; // array of iterators to runs
iterator _Mi; // middle iterator
iterator _Li; // last (end) iterator
size_t _I; // index to _Ai
for (_I = 0; _I < _ASZ; _I++) // "empty" array
_Ai[_I] = _Last; // _Ai[] == _Last => empty entry
// merge nodes into array
for (_Li = _First; _Li != _Last;) {
_Mi = _Li++;
for (_I = 0; (_I < _ASZ) && _Ai[_I] != _Last; _I++) {
_Mi = _Merge(_Pass_fn(_Pred), _Ai[_I], _Mi, _Li);
_Ai[_I] = _Last;
}
if (_I == _ASZ)
_I--;
_Ai[_I] = _Mi;
}
// merge array runs into single run
for (_I = 0; _I < _ASZ && _Ai[_I] == _Last; _I++);
_Mi = _Ai[_I++];
while (1) {
for (; _I < _ASZ && _Ai[_I] == _Last; _I++);
if (_I == _ASZ)
break;
_Mi = _Merge(_Pass_fn(_Pred), _Ai[_I++], _Mi, _Last);
}
}
这个答案的其余部分是历史性的.
基于@IgorTandetnik的演示，我能够重现该问题(旧排序无法编译，新版本可以正常工作):
#include <iostream>
#include <list>
#include <memory>
template <typename T>
class MyAlloc : public std::allocator<T> {
public:
MyAlloc(T) {} // suppress default constructor
template <typename U>
MyAlloc(const MyAlloc<U>& other) : std::allocator<T>(other) {}
template< class U > struct rebind { typedef MyAlloc<U> other; };
};
int main()
{
std::list<int, MyAlloc<int>> l(MyAlloc<int>(0));
l.push_back(3);
l.push_back(0);
l.push_back(2);
l.push_back(1);
l.sort();
return 0;
}
我在2016年7月注意到了这一变化，并在2016年8月1日通过电子邮件将P.J. Plauger告知了这一变化.
有趣的是，我们的更改日志未反映此更改.那 可能意味着它是由我们的一个较大客户建议"的， 在代码审查中由我得到.我现在所知道的就是变化来了 大约在2015年秋天 令我震惊的是那条线:
iterator _Mid = _STD next(_First, _Size / 2);对于一个大型列表，当然可能要花费很长的时间.
该代码看上去比我1995年初写的要优雅得多(！)， 但绝对会增加时间复杂度.该版本是仿照的 是继Stepanov，Lee和Musser在原始STL中采用的方法之后. 在选择算法时很少发现它们是错误的.
我现在正在恢复原始代码的最新已知良好版本.
我不知道P.J. Plauger还原到原始代码是否处理了新的分配器问题，或者Microsoft是否或如何与Dinkumware进行交互.
为了比较自上而下和自下而上的方法，我创建了一个包含400万个元素的链表，每个元素由一个64位无符号整数组成，假设我最终得到的是几乎按顺序排列的节点的双链表( (即使可以动态分配)，请用随机数填充它们，然后对其进行排序.节点不会移动，只会更改链接，但是现在遍历列表将以随机顺序访问节点.然后，我用另一组随机数填充这些随机排序的节点，然后再次对其进行排序.我将2015年自上而下的方法与先前的自下而上的方法进行了比较，以适应2015年所做的其他更改(sort()现在使用谓词compare函数而不是两个单独的函数调用sort()).这些就是结果. update -我添加了一个基于节点指针的版本，并且还指出了从列表中简单创建矢量，对矢量进行排序，复制回来的时间.
sequential nodes: 2015 version 1.6 seconds, prior version 1.5 seconds
random nodes: 2015 version 4.0 seconds, prior version 2.8 seconds
random nodes: node pointer based version 2.6 seconds
random nodes: create vector from list, sort, copy back 1.25 seconds
对于顺序节点，先前版本仅快一点，但对于随机节点，先前版本快30％，节点指针版本快35％，并从列表中创建向量，对向量进行排序，则复制速度提高了69％.
以下是std :: list :: sort()的第一个替换代码，我过去使用小数组(_BinList [])方法与使用VS2015的自上而下方法进行了先自下而上的比较，我希望这种比较是公平的，因此我修改了<列表>.
void sort()
{ // order sequence, using operator<
sort(less<>());
}
template<class _Pr2>
void sort(_Pr2 _Pred)
{ // order sequence, using _Pred
if (2 > this->_Mysize())
return;
const size_t _MAXBINS = 25;
_Myt _Templist, _Binlist[_MAXBINS];
while (!empty())
{
// _Templist = next element
_Templist._Splice_same(_Templist.begin(), *this, begin(),
++begin(), 1);
// merge with array of ever larger bins
size_t _Bin;
for (_Bin = 0; _Bin < _MAXBINS && !_Binlist[_Bin].empty();
++_Bin)
_Templist.merge(_Binlist[_Bin], _Pred);
// don't go past end of array
if (_Bin == _MAXBINS)
_Bin--;
// update bin with merged list, empty _Templist
_Binlist[_Bin].swap(_Templist);
}
// merge bins back into caller's list
for (size_t _Bin = 0; _Bin < _MAXBINS; _Bin++)
if(!_Binlist[_Bin].empty())
this->merge(_Binlist[_Bin], _Pred);
}
我做了一些小改动.原始代码在名为_Maxbin的变量中跟踪了实际的最大bin，但是最终合并中的开销很小，以至于我删除了与_Maxbin相关联的代码.在数组构建期间，原始代码的内部循环合并为_Binlist []元素，然后交换为_Templist，这似乎毫无意义.我将内部循环更改为仅合并到_Templist中，仅在找到空的_Binlist []元素后才进行交换.
下面是基于节点指针的std :: list :: sort()替代品，我用于另一个比较.这消除了与分配有关的问题.如果可能发生比较异常，则必须将阵列和临时列表(pNode)中的所有节点附加回原始列表，否则可能会将比较异常视为小于比较.
void sort()
{ // order sequence, using operator<
sort(less<>());
}
template<class _Pr2>
void sort(_Pr2 _Pred)
{ // order sequence, using _Pred
const size_t _NUMBINS = 25;
_Nodeptr aList[_NUMBINS]; // array of lists
_Nodeptr pNode;
_Nodeptr pNext;
_Nodeptr pPrev;
if (this->size() < 2) // return if nothing to do
return;
this->_Myhead()->_Prev->_Next = 0; // set last node ->_Next = 0
pNode = this->_Myhead()->_Next; // set ptr to start of list
size_t i;
for (i = 0; i < _NUMBINS; i++) // zero array
aList[i] = 0;
while (pNode != 0) // merge nodes into array
{
pNext = pNode->_Next;
pNode->_Next = 0;
for (i = 0; (i < _NUMBINS) && (aList[i] != 0); i++)
{
pNode = _MergeN(_Pred, aList[i], pNode);
aList[i] = 0;
}
if (i == _NUMBINS)
i--;
aList[i] = pNode;
pNode = pNext;
}
pNode = 0; // merge array into one list
for (i = 0; i < _NUMBINS; i++)
pNode = _MergeN(_Pred, aList[i], pNode);
this->_Myhead()->_Next = pNode; // update sentinel node links
pPrev = this->_Myhead(); // and _Prev pointers
while (pNode)
{
pNode->_Prev = pPrev;
pPrev = pNode;
pNode = pNode->_Next;
}
pPrev->_Next = this->_Myhead();
this->_Myhead()->_Prev = pPrev;
}
template<class _Pr2>
_Nodeptr _MergeN(_Pr2 &_Pred, _Nodeptr pSrc1, _Nodeptr pSrc2)
{
_Nodeptr pDst = 0; // destination head ptr
_Nodeptr *ppDst = &pDst; // ptr to head or prev->_Next
if (pSrc1 == 0)
return pSrc2;
if (pSrc2 == 0)
return pSrc1;
while (1)
{
if (_DEBUG_LT_PRED(_Pred, pSrc2->_Myval, pSrc1->_Myval))
{
*ppDst = pSrc2;
pSrc2 = *(ppDst = &pSrc2->_Next);
if (pSrc2 == 0)
{
*ppDst = pSrc1;
break;
}
}
else
{
*ppDst = pSrc1;
pSrc1 = *(ppDst = &pSrc1->_Next);
if (pSrc1 == 0)
{
*ppDst = pSrc2;
break;
}
}
}
return pDst;
}
I remember that since the beginning of times the most popular approach to implementing std::list<>::sort() was the classic Merge Sort algorithm implemented in bottom-up fashion (see also What makes the gcc std::list sort implementation so fast?).
I remember seeing someone aptly refer to this strategy as "onion chaining" approach.
At least that's the way it is in GCC's implementation of C++ standard library (see, for example, here). And this is how it was in old Dimkumware's STL in MSVC version of standard library, as well as in all versions of MSVC all the way to VS2013.
However, the standard library supplied with VS2015 suddenly no longer follows this sorting strategy. The library shipped with VS2015 uses a rather straightforward recursive implementation of top-down Merge Sort. This strikes me as strange, since top-down approach requires access to the mid-point of the list in order to split it in half. Since std::list<> does not support random access, the only way to find that mid-point is to literally iterate through half of the list. Also, at the very beginning it is necessary to know the total number of elements in the list (which was not necessarily an O(1) operation before C++11).
Nevertheless, std::list<>::sort() in VS2015 does exactly that. Here's an excerpt from that implementation that locates the mid-point and performs recursive calls
...
iterator _Mid = _STD next(_First, _Size / 2);
_First = _Sort(_First, _Mid, _Pred, _Size / 2);
_Mid = _Sort(_Mid, _Last, _Pred, _Size - _Size / 2);
...
As you can see, they just nonchalantly use std::next to walk through the first half of the list and arrive at _Mid iterator.
What could be the reason behind this switch, I wonder? All I see is a seemingly obvious inefficiency of repetitive calls to std::next at each level of recursion. Naive logic says that this is slower. If they are willing to pay this kind of price, they probably expect to get something in return. What are they getting then? I don't immediately see this algorithm as having better cache behavior (compared to the original bottom-up approach). I don't immediately see it as behaving better on pre-sorted sequences.
Granted, since C++11 std::list<> is basically required to store its element count, which makes the above slightly more efficient, since we always know the element count in advance. But that still does not seem to be enough to justify the sequential scan on each level of recursion.
(Admittedly, I haven't tried to race the implementations against each other. Maybe there are some surprises there.)
Note this answer has been updated to address all of the issues mentioned in the comments below and after the question, by making the same change from an array of lists to an array of iterators, while retaining the faster bottom up merge sort algorithm, and eliminating the small chance of stack overflow due to recursion with the top down merge sort algorithm.
The reason I didn't originally consider iterators was due to the VS2015 change to top down, leading me to believe there was some issue with trying to change the existing bottom up algorithm to use iterators. It was only when I tried to analyze the switch to iterators myself that I realized there was a solution.
In @sbi's comment, he asked the author of the top down approach, Stephan T. Lavavej, why the change was made. Stephan's response was "to avoid memory allocation and default constructing allocators". VS2015 introduced non-default-constructible and stateful allocators, which presents an issue when using the prior version's array of lists, as each instance of a list allocates a dummy node, and a change would be needed to handle no default allocator.
Lavavej's solution was to switch to using iterators to keep track of run boundaries within the original list instead of an internal array of lists. The merge logic was changed to use 3 iterator parameters, 1st parameter is iterator to start of left run, 2nd parameter is iterator to end of left run == iterator to start of right run, 3rd parameter is iterator to end of right run. The merge process uses std::list::splice to move nodes within the original list during merge operations. This has the added benefit of being exception safe. If a caller's compare function throws an exception, the list will be re-ordered, but no loss of data will occur (assuming splice can't fail). With the prior scheme, some (or most) of the data would be in the internal array of lists if an exception occurred, and data would be lost from the original list.
However the switch to top down merge sort was not needed. Initially, thinking there was some unknown to me reason for VS2015 switch to top down, I focused on using the internal interfaces in the same manner as std::list::splice. I later decided to investigate switching bottom up to use an array of iterators. I realized the order of runs stored in the internal array was newest (array[0] = rightmost) to oldest (array[last] = leftmost), and that it could use the same iterator based merge logic as VS2015's top down approach.
For bottom up merge sort, array[i] is an iterator to the start of a sorted sub-list with 2^i nodes, or it is empty (using std::list::end to indicate empty). The end of each sorted sub-list will be the start of a sorted sub-list in the next prior non-empty entry in the array, or if at the start of the array, in a local iterator (it points to end of newest run). Similar to the top down approach, the array of iterators is only used to keep track of sorted run boundaries within the original linked list, while the merge process uses std::list::splice to move nodes within the original linked list.
If a linked list is large and the nodes scattered, there will be a lot of cache misses. Bottom up will be about 30% faster than top down (equivalent to stating top down is about 42% slower than bottom up ). Then again, if there's enough memory, it would usually be faster to move the list to an array or vector, sort the array or vector, then create a new list from the sorted array or vector.
Example C++ code:
#define ASZ 32
template <typename T>
void SortList(std::list<T> &ll)
{
if (ll.size() < 2) // return if nothing to do
return;
std::list<T>::iterator ai[ASZ]; // array of iterators
std::list<T>::iterator mi; // middle iterator (end lft, bgn rgt)
std::list<T>::iterator ei; // end iterator
size_t i;
for (i = 0; i < ASZ; i++) // "clear" array
ai[i] = ll.end();
// merge nodes into array
for (ei = ll.begin(); ei != ll.end();) {
mi = ei++;
for (i = 0; (i < ASZ) && ai[i] != ll.end(); i++) {
mi = Merge(ll, ai[i], mi, ei);
ai[i] = ll.end();
}
if(i == ASZ)
i--;
ai[i] = mi;
}
// merge array into single list
ei = ll.end();
for(i = 0; (i < ASZ) && ai[i] == ei; i++);
mi = ai[i++];
while(1){
for( ; (i < ASZ) && ai[i] == ei; i++);
if (i == ASZ)
break;
mi = Merge(ll, ai[i++], mi, ei);
}
}
template <typename T>
typename std::list<T>::iterator Merge(std::list<T> &ll,
typename std::list<T>::iterator li,
typename std::list<T>::iterator mi,
typename std::list<T>::iterator ei)
{
std::list<T>::iterator ni;
(*mi < *li) ? ni = mi : ni = li;
while(1){
if(*mi < *li){
ll.splice(li, ll, mi++);
if(mi == ei)
return ni;
} else {
if(++li == mi)
return ni;
}
}
}
Example replacement code for VS2019's std::list::sort() (the merge logic was made into a separate internal function, since it's now used in two places).
private:
template <class _Pr2>
iterator _Merge(_Pr2 _Pred, iterator _First, iterator _Mid, iterator _Last){
iterator _Newfirst = _First;
for (bool _Initial_loop = true;;
_Initial_loop = false) { // [_First, _Mid) and [_Mid, _Last) are sorted and non-empty
if (_DEBUG_LT_PRED(_Pred, *_Mid, *_First)) { // consume _Mid
if (_Initial_loop) {
_Newfirst = _Mid; // update return value
}
splice(_First, *this, _Mid++);
if (_Mid == _Last) {
return _Newfirst; // exhausted [_Mid, _Last); done
}
}
else { // consume _First
++_First;
if (_First == _Mid) {
return _Newfirst; // exhausted [_First, _Mid); done
}
}
}
}
template <class _Pr2>
void _Sort(iterator _First, iterator _Last, _Pr2 _Pred,
size_type _Size) { // order [_First, _Last), using _Pred, return new first
// _Size must be distance from _First to _Last
if (_Size < 2) {
return; // nothing to do
}
const size_t _ASZ = 32; // array size
iterator _Ai[_ASZ]; // array of iterators to runs
iterator _Mi; // middle iterator
iterator _Li; // last (end) iterator
size_t _I; // index to _Ai
for (_I = 0; _I < _ASZ; _I++) // "empty" array
_Ai[_I] = _Last; // _Ai[] == _Last => empty entry
// merge nodes into array
for (_Li = _First; _Li != _Last;) {
_Mi = _Li++;
for (_I = 0; (_I < _ASZ) && _Ai[_I] != _Last; _I++) {
_Mi = _Merge(_Pass_fn(_Pred), _Ai[_I], _Mi, _Li);
_Ai[_I] = _Last;
}
if (_I == _ASZ)
_I--;
_Ai[_I] = _Mi;
}
// merge array runs into single run
for (_I = 0; _I < _ASZ && _Ai[_I] == _Last; _I++);
_Mi = _Ai[_I++];
while (1) {
for (; _I < _ASZ && _Ai[_I] == _Last; _I++);
if (_I == _ASZ)
break;
_Mi = _Merge(_Pass_fn(_Pred), _Ai[_I++], _Mi, _Last);
}
}
The remainder of this answer is historical.
I was able to reproduce the issue (old sort fails to compile, new one works) based on a demo from @IgorTandetnik:
#include <iostream>
#include <list>
#include <memory>
template <typename T>
class MyAlloc : public std::allocator<T> {
public:
MyAlloc(T) {} // suppress default constructor
template <typename U>
MyAlloc(const MyAlloc<U>& other) : std::allocator<T>(other) {}
template< class U > struct rebind { typedef MyAlloc<U> other; };
};
int main()
{
std::list<int, MyAlloc<int>> l(MyAlloc<int>(0));
l.push_back(3);
l.push_back(0);
l.push_back(2);
l.push_back(1);
l.sort();
return 0;
}
I noticed this change back in July, 2016 and emailed P.J. Plauger about this change on August 1, 2016. A snippet of his reply:
Interestingly enough, our change log doesn't reflect this change. That probably means it was "suggested" by one of our larger customers and got by me on the code review. All I know now is that the change came in around the autumn of 2015. When I reviewed the code, the first thing that struck me was the line:
iterator _Mid = _STD next(_First, _Size / 2);which, of course, can take a very long time for a large list.
The code looks a bit more elegant than what I wrote in early 1995(!), but definitely has worse time complexity. That version was modeled after the approach by Stepanov, Lee, and Musser in the original STL. They are seldom found to be wrong in their choice of algorithms.
I'm now reverting to our latest known good version of the original code.
I don't know if P.J. Plauger's reversion to the original code dealt with the new allocator issue, or if or how Microsoft interacts with Dinkumware.
For a comparison of the top down versus bottom up methods, I created a linked list with 4 million elements, each consisting of one 64 bit unsigned integer, assuming I would end up with a doubly linked list of nearly sequentially ordered nodes (even though they would be dynamically allocated), filled them with random numbers, then sorted them. The nodes don't move, only the linkage is changed, but now traversing the list accesses the nodes in random order. I then filled those randomly ordered nodes with another set of random numbers and sorted them again. I compared the 2015 top down approach with the prior bottom up approach modified to match the other changes made for 2015 (sort() now calls sort() with a predicate compare function, rather than having two separate functions). These are the results. update - I added a node pointer based version and also noted the time for simply creating a vector from list, sorting vector, copy back.
sequential nodes: 2015 version 1.6 seconds, prior version 1.5 seconds
random nodes: 2015 version 4.0 seconds, prior version 2.8 seconds
random nodes: node pointer based version 2.6 seconds
random nodes: create vector from list, sort, copy back 1.25 seconds
For sequential nodes, the prior version is only a bit faster, but for random nodes, the prior version is 30% faster, and the node pointer version 35% faster, and creating a vector from the list, sorting the vector, then copying back is 69% faster.
Below is the first replacement code for std::list::sort() I used to compare the prior bottom up with small array (_BinList[]) method versus VS2015's top down approach I wanted the comparison to be fair, so I modified a copy of < list >.
void sort()
{ // order sequence, using operator<
sort(less<>());
}
template<class _Pr2>
void sort(_Pr2 _Pred)
{ // order sequence, using _Pred
if (2 > this->_Mysize())
return;
const size_t _MAXBINS = 25;
_Myt _Templist, _Binlist[_MAXBINS];
while (!empty())
{
// _Templist = next element
_Templist._Splice_same(_Templist.begin(), *this, begin(),
++begin(), 1);
// merge with array of ever larger bins
size_t _Bin;
for (_Bin = 0; _Bin < _MAXBINS && !_Binlist[_Bin].empty();
++_Bin)
_Templist.merge(_Binlist[_Bin], _Pred);
// don't go past end of array
if (_Bin == _MAXBINS)
_Bin--;
// update bin with merged list, empty _Templist
_Binlist[_Bin].swap(_Templist);
}
// merge bins back into caller's list
for (size_t _Bin = 0; _Bin < _MAXBINS; _Bin++)
if(!_Binlist[_Bin].empty())
this->merge(_Binlist[_Bin], _Pred);
}
I made some minor changes. The original code kept track of the actual maximum bin in a variable named _Maxbin, but the overhead in the final merge is small enough that I removed the code associated with _Maxbin. During the array build, the original code's inner loop merged into a _Binlist[] element, followed by a swap into _Templist, which seemed pointless. I changed the inner loop to just merge into _Templist, only swapping once an empty _Binlist[] element is found.
Below is a node pointer based replacement for std::list::sort() I used for yet another comparison. This eliminates allocation related issues. If a compare exception is possible and occurred, all the nodes in the array and temp list (pNode) would have to be appended back to the original list, or possibly a compare exception could be treated as a less than compare.
void sort()
{ // order sequence, using operator<
sort(less<>());
}
template<class _Pr2>
void sort(_Pr2 _Pred)
{ // order sequence, using _Pred
const size_t _NUMBINS = 25;
_Nodeptr aList[_NUMBINS]; // array of lists
_Nodeptr pNode;
_Nodeptr pNext;
_Nodeptr pPrev;
if (this->size() < 2) // return if nothing to do
return;
this->_Myhead()->_Prev->_Next = 0; // set last node ->_Next = 0
pNode = this->_Myhead()->_Next; // set ptr to start of list
size_t i;
for (i = 0; i < _NUMBINS; i++) // zero array
aList[i] = 0;
while (pNode != 0) // merge nodes into array
{
pNext = pNode->_Next;
pNode->_Next = 0;
for (i = 0; (i < _NUMBINS) && (aList[i] != 0); i++)
{
pNode = _MergeN(_Pred, aList[i], pNode);
aList[i] = 0;
}
if (i == _NUMBINS)
i--;
aList[i] = pNode;
pNode = pNext;
}
pNode = 0; // merge array into one list
for (i = 0; i < _NUMBINS; i++)
pNode = _MergeN(_Pred, aList[i], pNode);
this->_Myhead()->_Next = pNode; // update sentinel node links
pPrev = this->_Myhead(); // and _Prev pointers
while (pNode)
{
pNode->_Prev = pPrev;
pPrev = pNode;
pNode = pNode->_Next;
}
pPrev->_Next = this->_Myhead();
this->_Myhead()->_Prev = pPrev;
}
template<class _Pr2>
_Nodeptr _MergeN(_Pr2 &_Pred, _Nodeptr pSrc1, _Nodeptr pSrc2)
{
_Nodeptr pDst = 0; // destination head ptr
_Nodeptr *ppDst = &pDst; // ptr to head or prev->_Next
if (pSrc1 == 0)
return pSrc2;
if (pSrc2 == 0)
return pSrc1;
while (1)
{
if (_DEBUG_LT_PRED(_Pred, pSrc2->_Myval, pSrc1->_Myval))
{
*ppDst = pSrc2;
pSrc2 = *(ppDst = &pSrc2->_Next);
if (pSrc2 == 0)
{
*ppDst = pSrc1;
break;
}
}
else
{
*ppDst = pSrc1;
pSrc1 = *(ppDst = &pSrc1->_Next);
if (pSrc1 == 0)
{
*ppDst = pSrc2;
break;
}
}
}
return pDst;
}
这篇关于`std :: list<> :: sort()`-为什么突然转向自上而下的策略?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
