我的代码分为一个部分在主线程中运行,另一部分在单独的线程中。主线程最后等待thread.join()
。
现在,如果主线程抛出异常,即使我尝试捕获异常,我的程序也会立即终止。我将其跟踪到std::~thread
,如果线程仍在运行(可连接),这将不令人满意。
在这种情况下,我希望异常能够等到线程自然退出后再继续传播。我怎么做?我是否需要用某种std::thread
来包装joining_thread
,将其加入?
#include <iostream>
#include <thread>
class foo
{
public:
foo(const char* name) : m_name(name) {}
~foo() { std::cout << "~foo - " << m_name << "\n"; }
private:
std::string m_name;
};
void doInThread()
{
std::this_thread::sleep_for(std::chrono::seconds(1));
std::cout << "doInThread\n";
}
int doit()
{
foo fooBefore("main before thread"); // d'tor is never called
auto thread = std::thread(doInThread);
foo fooAfter("main after thread");
// Do stuff, which might throw
throw std::exception("Something happened in main");
thread.join();
}
int main()
{
try
{
doit();
}
catch (std::exception ex)
{
std::cout << "Exception: " << ex.what() << "\n"; // Never happens - program terminates beforehand
}
}
您的程序通过调用std::terminate
函数中止。
当引发异常时,将发生堆栈展开,这意味着将删除在调用throw
之前创建的所有局部变量。当调用线程的dtor且线程处于可连接状态时,将调用std::terminate
。
使用RAII创建其dtor将加入您的线程的对象。
according to thread reference现在在展开堆栈时,将按以下方式调用析构函数:
template<class F> struct Cleaner { Cleaner(F in) : f(in) {} ~Cleaner() { f(); } F f; }; template<class F> Cleaner<F> makeCleaner(F f) { return Cleaner<F>(f); } int doit() { foo fooBefore("main before thread"); auto thread = std::thread(doInThread); auto cleaner = makeCleaner([&thread](){ thread.join(); }); foo fooAfter("main after thread"); throw std::runtime_error("Something happened in main"); // without throw statement, dtor of cleaner instance would be called here }
,fooAfter
(您在其中等待线程完成)和cleaner
。