QFuture内存泄漏

Question

我想并行化一个函数，问题是几个小时后，我的内存就过载了。

测试程序计算一些简单的东西，并且到目前为止还能工作。只有内存使用量在不断增加。

QT项目文件：

QT -= gui
QT += concurrent widgets
CONFIG += c++11 console
CONFIG -= app_bundle
DEFINES += QT_DEPRECATED_WARNINGS
SOURCES += main.cpp

QT程序文件：

#include <QCoreApplication>
#include <qdebug.h>
#include <qtconcurrentrun.h>

double parallel_function(int instance){
    return (double)(instance)*10.0;
}

int main(int argc, char *argv[])
{
    QCoreApplication a(argc, argv);
    int nr_of_threads = 8;
    double result_sum,temp_var;
    for(qint32 i = 0; i<100000000; i++){
      QFuture<double> * future = new QFuture<double>[nr_of_threads];

      for(int thread = 0; thread < nr_of_threads; thread++){
        future[thread] = QtConcurrent::run(parallel_function,thread);
      }
      for(int thread = 0; thread < nr_of_threads; thread++){
        future[thread].waitForFinished();
        temp_var = future[thread].result();
        qDebug()<<"result: " << temp_var;
        result_sum += temp_var;
      }
  }

  qDebug()<<"total: "<<result_sum;
  return a.exec();
}

正如我所观察到的，QtConcurrent::run(parallel_function,thread)分配内存，但在future[thread].waitForFinished()之后不释放内存。

这里怎么了？

Answer 1

内存泄漏是因为future数组未被删除。在外部for循环的末尾添加delete[] future。

for(qint32 i = 0; i<100000000; i++)
{
   QFuture<double> * future = new QFuture<double>[nr_of_threads];

   for(int thread = 0; thread < nr_of_threads; thread++){
     future[thread] = QtConcurrent::run(parallel_function,thread);
   }
   for(int thread = 0; thread < nr_of_threads; thread++){
      future[thread].waitForFinished();
      temp_var = future[thread].result();
      qDebug()<<"result: " << temp_var;
      result_sum += temp_var;
   }

   delete[] future; // <--
}

Answer 2

下面是它的外观--注意每件事情都可以变得简单得多！你已经下定决心要做手动内存管理了:为什么？首先，QFuture是一个值。您可以非常有效地将其存储在任何将为您管理内存的向量容器中。您可以使用range-for迭代这样的容器。等。

QT = concurrent   # dependencies are automatic, you don't use widgets
CONFIG += c++14 console
CONFIG -= app_bundle
SOURCES = main.cpp

尽管示例是合成的，而且map_function非常简单，但值得考虑的是如何以最高效和最有表现力的方式处理事情。您的算法是典型的map-还原操作，blockingMappedReduce的开销仅为手工完成所有工作的一半。

首先，让我们用C++重新定义原来的问题，而不是用一些C-加上Frankenstein。

// https://github.com/KubaO/stackoverflown/tree/master/questions/future-ranges-49107082
/* QtConcurrent will include QtCore as well */
#include <QtConcurrent>
#include <algorithm>
#include <iterator>

using result_type = double;

static result_type map_function(int instance){
   return instance * result_type(10);
}

static void sum_modifier(result_type &result, result_type value) {
   result += value;
}

static result_type sum_function(result_type result, result_type value) {
   return result + value;
}

result_type sum_approach1(int const N) {
   QVector<QFuture<result_type>> futures(N);
   int id = 0;
   for (auto &future : futures)
      future = QtConcurrent::run(map_function, id++);
   return std::accumulate(futures.cbegin(), futures.cend(), result_type{}, sum_function);
}

没有手动内存管理，也没有将线程显式拆分成“线程”--这是没有意义的，因为并发执行平台知道有多少线程。这已经更好了！

但这似乎相当浪费:每个未来内部分配至少一次(!)。

与其对每个结果显式地使用期货，我们还可以使用map-还原框架。为了生成序列，我们可以定义一个迭代器来提供我们想要处理的整数。迭代器可以是前向迭代器，也可以是双向迭代器，它的实现是QtConcurrent框架所需的最小值。

#include <iterator>

template <typename tag> class num_iterator : public std::iterator<tag, int, int, const int*, int> {
   int num = 0;
   using self = num_iterator;
   using base = std::iterator<tag, int, int, const int*, int>;
public:
   explicit num_iterator(int num = 0) : num(num) {}
   self &operator++() { num ++; return *this; }
   self &operator--() { num --; return *this; }
   self &operator+=(typename base::difference_type d) { num += d; return *this; }
   friend typename base::difference_type operator-(self lhs, self rhs) { return lhs.num - rhs.num; }
   bool operator==(self o) const { return num == o.num; }
   bool operator!=(self o) const { return !(*this == o); }
   typename base::reference operator*() const { return num; }
};

using num_f_iterator = num_iterator<std::forward_iterator_tag>;

result_type sum_approach2(int const N) {
   auto results = QtConcurrent::blockingMapped<QVector<result_type>>(num_f_iterator{0}, num_f_iterator{N}, map_function);
   return std::accumulate(results.cbegin(), results.cend(), result_type{}, sum_function);
}

using num_b_iterator = num_iterator<std::bidirectional_iterator_tag>;

result_type sum_approach3(int const N) {
   auto results = QtConcurrent::blockingMapped<QVector<result_type>>(num_b_iterator{0}, num_b_iterator{N}, map_function);
   return std::accumulate(results.cbegin(), results.cend(), result_type{}, sum_function);
}

我们可以放弃std::accumulate，改用blockingMappedReduced吗？当然：

result_type sum_approach4(int const N) {
   return QtConcurrent::blockingMappedReduced(num_b_iterator{0}, num_b_iterator{N},
                                              map_function, sum_modifier);
}

我们还可以尝试一个随机访问迭代器：

using num_r_iterator = num_iterator<std::random_access_iterator_tag>;

result_type sum_approach5(int const N) {
   return QtConcurrent::blockingMappedReduced(num_r_iterator{0}, num_r_iterator{N},
                                              map_function, sum_modifier);
}

最后，我们可以从使用范围生成迭代器切换到预先计算的范围：

#include <numeric>

result_type sum_approach6(int const N) {
   QVector<int> sequence(N);
   std::iota(sequence.begin(), sequence.end(), 0);
   return QtConcurrent::blockingMappedReduced(sequence, map_function, sum_modifier);
}

当然，我们的重点是对所有这些进行基准测试：

template <typename F> void benchmark(F fun, double const N) {
   QElapsedTimer timer;
   timer.start();
   auto result = fun(N);
   qDebug() << "sum:" << fixed << result << "took" << timer.elapsed()/N << "ms/item";
}

int main() {
   const int N = 1000000;
   benchmark(sum_approach1, N);
   benchmark(sum_approach2, N);
   benchmark(sum_approach3, N);
   benchmark(sum_approach4, N);
   benchmark(sum_approach5, N);
   benchmark(sum_approach6, N);
}

在我的系统上，在发布版本中，输出是：

sum: 4999995000000.000000 took 0.015778 ms/item
sum: 4999995000000.000000 took 0.003631 ms/item
sum: 4999995000000.000000 took 0.003610 ms/item
sum: 4999995000000.000000 took 0.005414 ms/item
sum: 4999995000000.000000 took 0.000011 ms/item
sum: 4999995000000.000000 took 0.000008 ms/item

注意如何在随机迭代序列上使用map-约简比使用QtConcurrent::run**，低3个数量级，比非随机迭代解快2个数量级。**