[英]Conditional std::future and std::async
I need to do conditional behavior. 我需要做条件行为。
std::future<int> f = pointer ? std::async(&Class::method, ptr) : 0;
// ... Some code
x = f.get();
So I would like to assign to x result async result of ptr->method()
call or 0 if ptr
is a nullptr
. 所以我想将
ptr->method()
调用的x结果异步结果分配给x,如果ptr
是nullptr
,则分配为0。
Is the code above ok? 上面的代码可以吗? Can I do anything like that (assign 'int' to 'std::futture'? Or maybe there is a better solution?
我可以这样做吗(将'int'分配给'std :: futture'?或者也许有更好的解决方案?
std::future
have no conversion constructor so your code is not valid (as you would have noticed if you actually tried to compile the code). std::future
没有转换构造函数,因此您的代码无效(如您实际上已尝试编译代码那样,您会注意到)。
What you can do is use a default-constructed future, and then check if it's valid before you use the future. 您可以使用默认构造的Future,然后在使用Future之前检查其是否有效 。
You can load a value into the future without using a thread like this: 您可以在不使用如下线程的情况下将值加载到将来:
std::future<int> f;
if ( pointer )
f = std::async(&Class::method, ptr);
else
{
std::promise<int> p;
p.set_value(0);
f = p.get_future();
}
// ... Some code
x = f.get();
But a simpler way to achieve the same goal would be: 但是,实现相同目标的一种更简单的方法是:
std::future<int> f;
if ( pointer )
f = std::async(&Class::method, ptr);
// ... Some code
x = f.valid() ? f.get() : 0;
You may return also a std::future
for your other case (using different policy): 您还可以为其他情况返回
std::future
(使用其他策略):
std::future<int> f = pointer
? std::async(&Class::method, ptr)
: std::async(std::launch::deferred, [](){ return 0;});
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