我正在尝试将局部变量移动到lambda的捕获中。
#include <thread>
#include <iostream>
// Moveable but not copyable object.
class WorkUnit
{
public:
WorkUnit(int) {}
WorkUnit(WorkUnit&&) noexcept {}
WorkUnit& operator=(WorkUnit&&) noexcept {return *this;}
WorkUnit(WorkUnit const&) = delete;
WorkUnit& operator=(WorkUnit const&) = delete;
// Non const function.
void doWork()
{
std::cerr << "Work\n";
}
};
int main()
{
WorkUnit data(4);
// Use C++14 generalized lambda capture.
std::thread test([data{std::move(data)}]()
{
// here it is complaining the `data` is a const value.
// Is there a way to capture this as a non const?
data.doWork();
}
);
test.join();
}
当我编译时,我得到了这个。
> g++ -std=c++14 WU.cpp
Test.cpp:26:13: error: member function 'doWork' not viable: 'this' argument has type 'const WorkUnit',
but function is not marked const
data.doWork();
^~~~
我希望捕获的值不是常量。
答案 0 :(得分:10)
您可以使用mutable:
mutable - 允许body修改copy复制的参数,并调用非const成员函数
除非在lambda表达式中使用关键字
mutable
,否则 function-call操作符是const限定的,是对象 通过副本捕获的内容不可从此operator()
内部修改。
std::thread test([data{std::move(data)}]() mutable
{
// the function-call operator is not const-qualified;
// then data is modifiable now
data.doWork();
}
);
值得注意的是,这允许修改由副本捕获的对象,这与原始对象无关。