我有一个程序,需要对大量的大数字分布进行排序。为了减少执行此操作所需的时间,我尝试对此进行多线程处理。
我为程序编写了一个简单的小摘要,以尝试找出问题所在。我相信我遇到了堆栈溢出或达到操作系统的堆栈限制,因为在以下情况下,我的测试程序会反射(reflect)段错误问题:
喵
#include <boost/thread/thread.hpp>
#include <vector>
#include <stdlib.h> // for rand()
void swapvals(double *distribution, const size_t &d1, const size_t &d2)
{
double temp = 0;
temp = distribution[d2];
distribution[d2] = distribution[d1];
distribution[d1] = temp;
//std::swap(distribution[d1], distribution[d2]);
}
size_t partition(double *distribution, size_t left, size_t right)
{
const double pivot = distribution[right];
while (left < right) {
while ((left < right) && distribution[left] <= pivot)
left++;
while ((left < right) && distribution[right] > pivot)
right--;
if (left < right)
{
swapvals(distribution, left, right);
}
}
return right;
}
void quickSort(double *distribution, const size_t left, const size_t right)
{
if (left >= right) {
return;
}
size_t part = partition(distribution, left, right);
quickSort(distribution, left, part - 1);
quickSort(distribution, part + 1, right);
}
void processDistribution(double *distributions, const size_t distribution_size)
{
std::clog << "beginning qsorting." << std::endl;
quickSort(distributions, 0, distribution_size - 1);
std::clog << "done qsorting." << std::endl;
}
int main(int argc, char* argv[])
{
size_t distribution_size = 65000;
size_t num_distributions = 10;
std::vector<double *> distributions;
// Create num_distributions distributions.
for (int i = 0; i < num_distributions; i++)
{
double * new_dist = new double[distribution_size];
for (int k = 0; k < distribution_size; k++)
{
// Works when I have actual numbers in the distributions.
// Seg faults when all the numbers are the same.
new_dist[k] =1;
//new_dist[k] = rand() % 1000 + 1; // uncomment this, and it works.
}
distributions.push_back(new_dist);
}
// Submit each distribution to a quicksort thread.
boost::thread_group threads;
for (std::vector<double *>::const_iterator it=distributions.begin(); it != distributions.end(); ++it)
{
// It works when I run processDistribution directly. Segfaults when I run it via threads.
//processDistribution(*it, distribution_size);
threads.create_thread(boost::bind(&processDistribution, *it, distribution_size));
}
threads.join_all();
// Show the results of the sort for all the distributions.
for (std::vector<double *>::const_iterator it=distributions.begin(); it != distributions.end(); ++it)
{
for (size_t i = 0; i < distribution_size; i++)
{
// print first and last 20 results.
if (i < 20 || i > (distribution_size - 20))
std::cout << (*it)[i] << ",";
}
std::cout << std::endl;
}
}
GDB对核心文件的分析得出:
Error in re-setting breakpoint -1: aix-thread: ptrace (52, 18220265) returned -1 (errno = 3 The process does not exist.)
Error in re-setting breakpoint -1: aix-thread: ptrace (52, 18220265) returned -1 (errno = 3 The process does not exist.)
Error in re-setting breakpoint -2: aix-thread: ptrace (52, 18220265) returned -1 (errno = 3 The process does not exist.)
Error in re-setting breakpoint -3: aix-thread: ptrace (52, 18220265) returned -1 (errno = 3 The process does not exist.)
Core was generated by `testthreads'.
Program terminated with signal SIGSEGV, Segmentation fault.
#0 0x00000001000056bc in partition (distribution=0x1101d1430, left=0, right=63626) at testthreads.cpp:18
warning: Source file is more recent than executable.
18
(gdb) bt 7
#0 0x00000001000056bc in partition (distribution=0x1101d1430, left=0, right=63626) at testthreads.cpp:18
#1 0x0000000100005834 in quickSort (distribution=0x1101d1430, left=0, right=63626) at testthreads.cpp:42
#2 0x0000000100005850 in quickSort (distribution=0x1101d1430, left=0, right=63627) at testthreads.cpp:43
#3 0x0000000100005850 in quickSort (distribution=0x1101d1430, left=0, right=63628) at testthreads.cpp:43
#4 0x0000000100005850 in quickSort (distribution=0x1101d1430, left=0, right=63629) at testthreads.cpp:43
#5 0x0000000100005850 in quickSort (distribution=0x1101d1430, left=0, right=63630) at testthreads.cpp:43
#6 0x0000000100005850 in quickSort (distribution=0x1101d1430, left=0, right=63631) at testthreads.cpp:43
(More stack frames follow...)
(gdb) frame 0
#0 0x00000001000056bc in partition (distribution=0x1101d1430, left=0, right=63626) at testthreads.cpp:18
18
(gdb) info locals
pivot = 1
(gdb) info args
distribution = 0x1101d1430
left = 0
right = 63626
(gdb)
另外,我的实际程序处理更多的线程和发行版。而且那里的GDB检查通常会显示更多看起来像内存损坏的异常堆栈跟踪(请注意,如何使用d1 = 12119调用swapVals,但在分区堆栈框架内将其作为4568618016来通过):
(gdb) bt 3
#0 0x00000001002aa0b8 in ScenRankReplacer<double>::swapvals (this=0xfffffffffffdfc8, distribution=..., d1=@0x1104c8178: 4568618016, d2=@0x1104c8140: 4568416720, ranking_values=0x1104c81d0,
r1=@0x1104c8170: 1152921504606838728, r2=@0x1002a16c8: 6917529029728344952) at ScenRankReplacer.h:96
#1 0x00000001002a7120 in ScenRankReplacer<double>::partition (this=0xfffffffffffdfc8, distribution=..., ranking_values=0x11069ae50, left=1, right=24237) at ScenRankReplacer.h:122
#2 0x00000001002a16c8 in ScenRankReplacer<double>::quickSort (this=0xfffffffffffdfc8, distribution=..., ranking_values=0x11069ae50, left=1, right=24237) at ScenRankReplacer.h:91
(More stack frames follow...)
(gdb) frame 1
#1 0x00000001002a7120 in ScenRankReplacer<double>::partition (this=0xfffffffffffdfc8, distribution=..., ranking_values=0x11069ae50, left=1, right=24237) at ScenRankReplacer.h:122
122 swapvals(distribution, mid, left, ranking_values, mid - 1, left - 1);
(gdb) p mid
$1 = 12119
(gdb) p left
$2 = 1
所以...我的问题:
堆栈限制?
大规模数据集上的递归快速排序不可行吗?
编译级别为O2时发生错误。
线程模型:aix
gcc版本4.8.3(GCC)
最佳答案
看起来可能与堆栈空间有关。线程很重要,因为尽管所有线程都有自己的堆栈,但是这些堆栈都共享相同的内存池。堆栈通常会根据需要增长,直到它们运行到已使用的内存中为止,在这种情况下,该内存很可能是来自另一个线程的堆栈。单线程程序不会有这个问题,并且可以使它的堆栈更大。 (另外,对于多个线程,您同时执行多种排序,这将需要更多的堆栈空间。)
解决此问题的一种方法是删除递归,并使用一些循环和本地存储来替换它。像这样的(未编译或经过测试)的代码:
void quickSort(double *distribution, size_t left, size_t right) {
std::vector<std::pair<size_t, size_t>> ranges;
for (;;) {
for (;;) {
if (left <= right)
break;
size_t part = partition(distribution, left, right);
// save range for later to replace the second recursive call
ranges.push_back(std::make_pair(part + 1, right));
// set right == part - 1, then loop, to replace the first recursive call
right = part - 1;
}
if (ranges.empty())
break;
// Take top off of ranges for the next loop, replacing the second recursive call
left = ranges.back().first;
right = ranges.back().second;
ranges.pop_back();
}
}
关于c++ - C++/调试(在AIX上为g++)导致段错误的递归Quicksort,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/32575470/