问题描述
每当方便时，我都在尝试通过使用异步来在程序中获得更好的性能.我的程序可以编译，但是每次使用包含异步调用的函数时，都会出现以下错误:
I'm trying to obtain a better performance in my program by using async whenever this is convenient. My program compiles, but I get the following error every time I use a function containing async calls:
C++ exception with description "No associated state"
我尝试使用lambda调用异步的方式例如如下:
The way I am trying to call async with a lambda is e.g. as follows:
auto f = [this](const Cursor& c){ return this->getAbsIndex(c); };
auto nodeAbsIndex = std::async(f,node); // node is const Cursor&
auto otherAbsIndex = std::async(f,other); // other too
size_t from = std::min(nodeAbsIndex.get(), otherAbsIndex.get());
size_t to = std::max(nodeAbsIndex.get(), otherAbsIndex.get());
要调用的函数的签名如下:
Signature of the function to call is as follows:
uint64_t getAbsIndex(const Cursor& c) const
我在这里做错了什么?感谢您的提示！迭戈
What am I doing wrong here? Thanks for any hints!Diego
推荐答案
您不能在同一将来两次致电get().仔细阅读文档(与valid()有关的部分): http://en.cppreference .com/w/cpp/thread/future/get
You can't call get() twice on the same future. Read documentation carefully (the part regarding valid()): http://en.cppreference.com/w/cpp/thread/future/get
另一方面，将uint64_t隐式转换为size_t是不好的.后者的尺寸可能较小.
On a side note, implicitly casting uint64_t to size_t is not good. The latter could be of smaller size.
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