使用tbb：：tbb::parallel_for中排队互斥的简单示例程序不编译

Question

这里是一个玩具例子，我正在玩，以学习如何使用TBB。并行：：operator()应该并行运行，但是它有一个关键区域，一次只能由一个处理器访问，所以它打印的消息不会被加扰。我的问题是，它无法编译，编译器消息对我没有多大帮助。我做错了什么？

此外，这是否被认为是在parallel_for中实现互斥的正确方法？

#include <iostream>
#include <vector>
#include <cmath>
#include <tbb/tbb.h>

typedef tbb::queuing_mutex Mutex;

struct Parallel
{
    Mutex mutex;
    std::vector<int> * values;

    Parallel(std::vector<int> * values_) : values(values_) {}

    void operator()( tbb::blocked_range< unsigned int > & range ) const {
        for(unsigned int i = range.begin(); i < range.end(); ++i) {
            {
                Mutex::scoped_lock lock(mutex);
                if ( (*values)[i] > 40)
                {
                    std::cout << "NO SCRAMBLING ALLOWED!\n";
                    std::cout.flush();
                }
                lock.release();
            }
        }
    }
};

int main() {
    const int someValue = 20000;

    std::vector<int> data(someValue);
    for(int i = 0; i < someValue; ++i) {
        data[i] = std::rand();
    }

    tbb::parallel_for( tbb::blocked_range<unsigned int>(0, data.size()),
                       Parallel(&data) );
}

Bellow是错误消息：

/path-to-src/main.cpp: In member function 'void Parallel::operator()(tbb::blocked_range<unsigned int>&) const':
/path-to-src/main.cpp:20:46: error: no matching function for call to 'tbb::queuing_mutex::scoped_lock::scoped_lock(const Mutex&)'
/path-to-src/main.cpp:20:46: note: candidates are:
In file included from /usr/include/tbb/tbb.h:65:0,
                 from /path-to-src/main.cpp:4:
/usr/include/tbb/queuing_mutex.h:80:9: note: tbb::queuing_mutex::scoped_lock::scoped_lock(tbb::queuing_mutex&)
/usr/include/tbb/queuing_mutex.h:80:9: note:   no known conversion for argument 1 from 'const Mutex {aka const tbb::queuing_mutex}' to 'tbb::queuing_mutex&'
/usr/include/tbb/queuing_mutex.h:77:9: note: tbb::queuing_mutex::scoped_lock::scoped_lock()
/usr/include/tbb/queuing_mutex.h:77:9: note:   candidate expects 0 arguments, 1 provided
/usr/include/tbb/queuing_mutex.h:66:11: note: tbb::queuing_mutex::scoped_lock::scoped_lock(const tbb::queuing_mutex::scoped_lock&)
/usr/include/tbb/queuing_mutex.h:66:11: note:   no known conversion for argument 1 from 'const Mutex {aka const tbb::queuing_mutex}' to 'const tbb::queuing_mutex::scoped_lock&'

Answer 1

tbb::parallel_for的设计目的是不按编写的方式编译示例。这是为了防止代码中的漏洞。tbb::parallel_for的范围形式通过值将函子复制到多个任务对象中。因此，每个任务都将有一个互斥体的单独副本，因此互斥体不会提供预期的同步。

修复代码的方法是在struct并行的外部声明互斥体，并通过指针传递它，类似于“值”的指针。

Answer 2

一切都是正确的，除了mutex必须是mutable才能在锁中使用。此外，释放锁是不必要的。

typedef tbb::queuing_mutex Mutex;

struct Parallel
{
    mutable Mutex mutex;              // mutable so that we can use it in const member
    std::vector<int> * values;

    Parallel(std::vector<int> * values_) : values(values_) {}

    // note: this is a const member
    void operator()( tbb::blocked_range< unsigned int > & range ) const
    {
        for(unsigned int i = range.begin(); i < range.end(); ++i)
        {
            Mutex::scoped_lock lock(mutex);       // requires non-const argument
            if ( (*values)[i] > 40)
            {
                std::cout << "NO SCRAMBLING ALLOWED!\n";
                std::cout.flush();
            }
            // no need to release the lock: the destructor will do that for you
        }
    }
};

当然，您也可以将成员函数设置为非const。