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C++/调试(AIX 上的 g++)递归快速排序导致段错误

转载 作者:塔克拉玛干 更新时间:2023-11-03 07:16:47 24 4
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我有一个程序需要对大量的大数值分布进行排序。为了减少执行此操作所需的时间,我正在尝试对其进行多线程处理。

我写了一个小而简单的程序抽象来尝试隔离问题。我相信我遇到了堆栈溢出,或者达到了操作系统的堆栈限制,因为我的测试程序在以下情况下反射(reflect)了段错误问题:

  • 分布都是相同的值(这意味着 qsort 会像废话一样运行)
  • 已启用线程。

喵喵

#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;
}

}

核心文件产量的 GDB 分析:

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)

此外,我的实际程序处理更多的线程和分布。那里的 GDB 检查经常显示更奇怪的堆栈跟踪,看起来像内存损坏(注意 swapVals 是如何用 d1 = 12119 调用的,但在分区堆栈框架内它作为 4568618016 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:

  1. 我说得对吗?我达到了堆栈限制吗?
  2. 我究竟如何确定情况确实如此(除了我在上面所做的推论之外)?有没有简单的方法来检测这些? GDB 线索之类的?
  3. 为什么线程很重要?所有的线程共享相同的吗堆栈限制?
  4. 最重要的是:我如何让它工作?!是一个对海量数据集进行递归快速排序不可行?

错误发生在编译级别O2。线程模型:aixgcc 版本 4.8.3 (GCC)

最佳答案

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,但这些堆栈都共享同一个内存池。堆栈通常会根据需要增长,直到它们遇到已使用的内存,在这种情况下,这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,并且可以增加它的堆栈。 (对于多个线程,您同时进行多种排序,这将需要更多堆栈空间。)

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(未编译或测试过的)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++/调试(AIX 上的 g++)递归快速排序导致段错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32575470/

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