我有一个程序，我需要对大量的大型数字发行版进行排序。为了减少执行此操作所需的时间，我正在尝试多线程。
我写了一个小的，简单的程序抽象来试图找出问题。我相信我遇到堆栈溢出或达到操作系统的堆栈限制，因为我的测试程序反映了分段错误问题：
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;
}
}
核心文件的GDB分析产生：
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。那里的GDB检查通常会显示更多奇怪的堆栈跟踪，看起来像内存损坏（请注意如何使用d1 = 12119调用swapVals，但是在分区堆栈框架内它通过4568618016进行调用）：
(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：
编译级别O2发生错误。 线程模型：aix gcc版本4.8.3（GCC）
答案 0 :(得分：1)
这看起来可能与堆栈空间有关。线程很重要，因为虽然所有线程都有自己的堆栈，但这些堆栈都共享相同的内存池。堆栈通常会根据需要增长，直到它们进入已经使用的内存，在这种情况下可能是来自另一个线程的堆栈。单线程程序不会出现这个问题，并且可以使堆栈更大。 （同时使用多个线程，您需要同时进行多种排序，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（未编译或测试）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();
}
}
答案 1 :(得分：0)
所以经过一点点拉毛后，我已经找到了问题的答案。
我说错了吗？我是否达到了堆叠限制？我到底怎么样？ 确定是这种情况（除了我所做的扣除之外） 以上）？ AND
有没有简单的方法来检测这些？ GDB线索或 什么？
A：是的。程序溢出了堆栈。我无法确定直接的方法来确定AIX上的情况。但是，当我将代码放入Windows上的Visual Studio 2015并运行它时，程序崩溃时出现明显的“Stack Overflow”错误。
我希望有一种方法可以在AIX上获得明确的“Stack Overflow”错误，类似于VS结果。我找不到办法。即使使用-fstack-check编译也没有给我一个存储错误:(
答： AIX上线程的默认堆栈大小非常小！
From this IBM developerworks blog post:
对于AIX上的32位编译应用程序，默认的pthread stacksize为96 KB;对于AIX上的64位编译应用程序，
我只能想到两种方式： A1：第一个是增加堆栈大小。
From the IBM Debugging Guidelines for Threads 线程的最小堆栈大小为96KB。它也是默认的堆栈大小。可以使用pthread.h头文件中定义的PTHREAD_STACK_MIN符号常量在编译时检索此编号。
请注意，最大堆栈大小为256MB，即段的大小。此限制由pthread.h头文件中的PTHREAD_STACK_MAX符号常量指示。
因此可以将堆栈大小增加到最大256MB，这是相当多的。
A2：另一种方法是简单地避免可能未绑定的递归。我的数据集非常大。可能不够大，无法花费256MB的堆栈，但迭代地重写quicksort函数是相当简单的。
void quickSort_iter(double *distribution, size_t left, size_t right)
{
if (left >= right)
return;
std::stack<std::pair<size_t, size_t> > partition_stack;
partition_stack.push(std::pair<size_t, size_t>(left, right));
while (!partition_stack.empty())
{
left = partition_stack.top().first;
right = partition_stack.top().second;
partition_stack.pop();
size_t pivot = partition(distribution, left, right);
if (pivot > 1)
partition_stack.push(std::pair<size_t, size_t>(left, pivot - 1));
if (pivot + 1 < right)
partition_stack.push(std::pair<size_t, size_t>(pivot + 1, right));
}
}