我正在尝试为嵌入式设备编译以下代码(它是TI的交叉编译器,具有C ++ 11(C ++ 0)的实验支持)。目标:
arm-arago-linux-gnueabi线程模型:posix gcc版本4.5.3 20110311 (预发布)(GCC)
无法编译移动构造函数和移动分配运算符的默认说明符(/home/user/test/main.cpp:40:26: error: 'th& th::operator=(th&&)' cannot be defaulted
)。
std::make_unique
& emplace_back
未实现,那些不可用。
我需要更改代码才能使其适用于此平台?
class th {
public:
void func() {
sleep(3);
*this->progress = 100;
}
th(int* prog) :
progress(prog),
m_thread(std::thread(&th::func, this)) {};
th(th const& other) = delete;
th(th && other) = default;
th& operator=(th const& other) = delete;
th& operator=(th &&) = default;
void join() { m_thread.join(); }
int *progress;
private:
std::thread m_thread;
};
int main(void) {
std::vector<int> progress;
progress.push_back(-1);
progress.push_back(-1);
std::deque<std::unique_ptr<th>> deq;
std::cout << "progress[0]:" << progress[0] << std::endl;
std::cout << "progress[1]:" << progress[1] << std::endl;
std::cout << "executing threads..." << std::endl;
for(size_t i = 0; i < 2; ++i) {
deq.push_back(std::unique_ptr<th>(new th(&progress[i])));
}
while(true) {
std::cout << "SIZE:" << deq.size() << std::endl;
if(deq.size() == 0)
break;
for (std::deque<std::unique_ptr<th>>::iterator it = deq.begin(); it != deq.end(); it++) {
//std::cout << (*it)->progress << std::endl;
if(*((*it)->progress) == 100) {
std::cout << "JOIN & DELETE" << std::endl;
(*it)->join();
deq.erase(it);
}
else {
std::cout << "STILL RUNNING" << std::endl;
}
//std::cout << *((*it)->progress) << std::endl;
}
sleep(1);
}
exit(EXIT_SUCCESS);
}
答案 0 :(得分:1)
gcc 4.5.x不支持为gcc C ++ 11支持功能生成移动特殊成员函数see N3053 for more details和https://gcc.gnu.org/gcc-4.5/cxx0x_status.html,所以你运气不好编译器。您的代码在gcc5 / 6和clang上编译正常,请查看它here。
一种选择是完全删除行
th(th const& other) = delete;
th(th && other) = default;
th& operator=(th const& other) = delete;
th& operator=(th &&) = default;
让编译器完成工作并为您生成特殊的成员函数。顺便说一句,默认情况下将删除复制ctor和复制赋值运算符,因为std::thread
是不可复制的。