我已经做了一些谷歌搜索，并在堆栈溢出方面看到了类似的问题，但是我无法理解处理它的原因/解决方案。鉴于在ThreadPool.hpp中声明的以下类，我收到以下2个错误：
错误1错误C2248：'std :: mutex :: mutex'：无法访问在类'std :: mutex'中声明的私有成员c：\ users \ jesse \ documents \ school \ summer semester 2012 \ concurrent processing \ project 2 \ ultra_grep v2 \ ultra_grep \ threadpool.hpp 39 1 ultra_grep2
错误2错误C2248：'std :: condition_variable :: condition_variable'：无法访问在类'std :: condition_variable'中声明的私有成员c：\ users \ jesse \ documents \ school \ summer semester 2012 \ concurrent processing \ project 2 \ ultra_grep v2 \ ultra_grep \ threadpool.hpp 39 1 ultra_grep2
class ThreadPool
{
private:
std::queue<std::string> _consoleTasks;
std::queue<std::tr2::sys::path> _tasks;
std::map<std::string, std::vector<GrepMatch>> _grepMatches;
int _nThreads, _fileMatches, _totalMatches, _workingThreads;
std::vector<thread> _threads;
Arguments _args;
std::mutex _taskMutex, _wakeMutex, _consoleMutex, _threadCountMutex;
std::condition_variable _wakeCondition;
public:
ThreadPool( int threads, Arguments args );
~ThreadPool() {};
void GrepFunc();
void ConsoleFunc();
void SearchFile( std::string );
void SearchFileVerbose( std::string );
void DisplayGrepResults();
queue<std::tr2::sys::path> Tasks() { return _tasks; }
};
以及在ThreadPool.cpp中实现的代码：
#include "ThreadPool.hpp"
using namespace std;
namespace fs = std::tr2::sys;
ThreadPool::ThreadPool( int threads, Arguments args )
: _nThreads( threads ), _args( args ), _fileMatches( 0 ), _totalMatches( 0 ), _workingThreads( 0 )
{
for( int i = 0; i < _nThreads; ++i )
{
_threads.push_back( thread( &ThreadPool::GrepFunc, this ) );
_tasks.push( args.root() );
++_workingThreads;
_wakeCondition.notify_one();
}
for( auto& t : _threads )
{
t.join();
}
DisplayGrepResults();
}
void ThreadPool::GrepFunc()
{
// implement a barrier()
while( !_workingThreads )
{
{ unique_lock<mutex> lk( _wakeMutex );
_wakeCondition.wait( lk ); }
while( !_tasks.empty() )
{
fs::path task;
bool gotTask = false;
{
lock_guard<mutex> tl( _taskMutex );
if( !_tasks.empty() )
{
{ lock_guard<mutex> tc( _threadCountMutex );
++_workingThreads; }
task = _tasks.front();
_tasks.pop();
gotTask = true;
}
}
if( gotTask )
{
if( fs::is_directory( task ) )
{
for( fs::directory_iterator dirIter( task ), endIter; dirIter != endIter; ++dirIter )
{
if( fs::is_directory( dirIter->path() ) )
{
{ lock_guard<mutex> tl( _taskMutex );
_tasks.push( dirIter->path() ); }
_wakeCondition.notify_one();
}
else
{
for( auto& e : _args.extensions() )
{
if( !dirIter->path().extension().compare( e ) )
{
{ lock_guard<mutex> tl( _taskMutex );
_tasks.push( dirIter->path() ); }
_wakeCondition.notify_one();
//SearchFile( dirIter->path() );
}
}
}
}
}
else
{
for( auto& e : _args.extensions() )
{
if( !task.extension().compare( e ) )
{
if( _args.is_verbose() )
SearchFile( task );
else
SearchFileVerbose( task );
}
}
}
{ lock_guard<mutex> tc( _threadCountMutex) ;
--_workingThreads; }
}
}
}
}
void ThreadPool::SearchFileVerbose( string path )
{
fstream file;
file.open( path );
if( !file )
{
// error handling
}
else
{
{
lock_guard<mutex> cm( _consoleMutex );
cout << "\nGrepping: " << path << endl;
int lineNumber = 1;
string line;
vector<GrepMatch> matches;
while( getline( file, line ) )
{
int lineMatches = 0;
sregex_token_iterator end;
for (sregex_token_iterator i(line.cbegin(), line.cend(), _args.regular_expression() );
i != end;
++i)
{
++lineMatches;
}
if( lineMatches > 0 )
{
GrepMatch match = GrepMatch( lineNumber, lineMatches, line );
matches.push_back( match );
cout << "Matched " << lineMatches << ": " << path << " [" << lineNumber << "] " << line << endl;
}
++lineNumber;
}
if( !matches.empty() )
{
_grepMatches[ path ] = matches;
}
}
}
}
我在互斥锁和condition_variable上都遇到错误，说他们无法访问在类中声明的私有成员。我的理解是，这是指复制构造函数？虽然我不明白为什么会发挥作用，因为我不知道我在哪里制作副本，因为他们只是班上的私人成员。
根据初始注释，我只实例化一个ThreadPool实例，并且不会在任何地方复制。我试图触摸互斥锁的唯一一次是在我班级的.cpp实现中。
任何人都可以帮助我更好地理解为什么会这样吗？
对于那些感兴趣的人是编译器输出：
1>------ Build started: Project: ultra_grep2, Configuration: Debug Win32 ------
1> ultra_grep_main.cpp
1> Unknown compiler version - please run the configure tests and report the results
1>c:\users\jesse\documents\school\summer semester 2012\concurrent processing\project 2\ultra_grep v2\ultra_grep\threadpool.hpp(39): error C2248: 'std::mutex::mutex' : cannot access private member declared in class 'std::mutex'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(116) : see declaration of 'std::mutex::mutex'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(107) : see declaration of 'std::mutex'
1> This diagnostic occurred in the compiler generated function 'ThreadPool::ThreadPool(const ThreadPool &)'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(116) : see declaration of 'std::mutex::mutex'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(107) : see declaration of 'std::mutex'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(116) : see declaration of 'std::mutex::mutex'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(107) : see declaration of 'std::mutex'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(116) : see declaration of 'std::mutex::mutex'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\mutex(107) : see declaration of 'std::mutex'
1>c:\users\jesse\documents\school\summer semester 2012\concurrent processing\project 2\ultra_grep v2\ultra_grep\threadpool.hpp(39): error C2248: 'std::condition_variable::condition_variable' : cannot access private member declared in class 'std::condition_variable'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\condition_variable(45) : see declaration of 'std::condition_variable::condition_variable'
1> c:\program files (x86)\microsoft visual studio 11.0\vc\include\condition_variable(30) : see declaration of 'std::condition_variable'
1> This diagnostic occurred in the compiler generated function 'ThreadPool::ThreadPool(const ThreadPool &)'
1> ThreadPool.cpp
1> Generating Code...
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
答案 0 :(得分：7)
问题是你的想法;编译器生成的复制构造函数和赋值运算符失败，因为mutex和condition_variable不可复制。在许多情况下，编译器仍会为这两者生成代码，即使您没有故意调用它们也不容易看到。根据我的经验，这个错误通常是由于尝试将该类型与标准库容器一起使用而引起的。
你应该添加一个空的私有拷贝构造函数和赋值运算符，即使你没有这个问题;我认为尝试复制线程池是不可取的。
答案 1 :(得分：3)
错误出现在：
1＆GT; ultra_grep_main.cpp
ThreadPool的复制构造函数使用：
此诊断发生在编译器生成的函数'ThreadPool :: ThreadPool（const ThreadPool＆amp;）'
我真的很惊讶编译器没有指出源文件中的哪一行隐式生成了复制构造函数。
在评论中，您询问复制构造函数应该如何。一个更好的问题是复制ThreadPool对象是否有意义，答案是一般情况下不会。要改进错误消息，您可以将复制构造函数标记为deleted，这样（希望）编译器将生成更好的错误报告，指出需要复制的位置：
class ThreadPool {
ThreadPool( ThreadPool const & ) = delete;
// ...
};
如果您的编译器不支持此C ++ 11功能，则替代方法是声明复制构造函数，但不定义它。这将在复制构造函数的使用位置触发访问错误。