[英]std::unique_ptr with std::map
I have a std::map
where the key is std::shared_ptr<Foo>
and the value is std::unique_ptr<Bar>
where Foo
and Bar
are very different classes from a third-party library.我有一个
std::map
,其中键是std::shared_ptr<Foo>
,值是std::unique_ptr<Bar>
,其中Foo
和Bar
是来自第三方库的非常不同的类。 I am using this std::map
object as an in-memory cache.我使用这个
std::map
对象作为内存缓存。
I am wondering what the best way of inserting a new entry into this map will be and then returned from a method, given that the Bar
passed into the std::unique_ptr
will already be constructed?我想知道在此映射中插入新条目然后从方法返回的最佳方法是什么,因为传入
std::unique_ptr
的Bar
已经被构造?
I currently have the following:我目前有以下几点:
class SomeClass
{
public:
const Bar* TryGetBarValue(std::shared_ptr<Foo> foo)
{
auto it = _cache.find(foo);
if(it == _cache.end())
{
Bar bar = ThirdPartLibrary::CreateBar();
_cache.emplace(foo, std::make_unique<Bar>(bar));
return _cache.rbegin()->second.get();
}
//return result as raw ptr from unique_ptr
return it->second.get();
}
private:
std::map<std::shared_ptr<Foo>, std::unique_ptr<Bar>> _cache;
}
EDIT编辑
Thanks to the answer provided by Quentin, this is now my implementation:感谢 Quentin 提供的答案,现在这是我的实现:
class SomeClass
{
public:
const Bar* TryGetBarValue(std::shared_ptr<Foo> foo)
{
auto it = _cachedImages.find(texture);
if (it != _cachedImages.end())
{
return it->second.get();
}
return _cachedImages.emplace(
std::move(texture),
std::make_unique<sf::Image>(texture->copyToImage())
).first->second.get();
}
private:
std::map<std::shared_ptr<Foo>, std::unique_ptr<Bar>> _cache;
}
Thanks for all your help!感谢你的帮助!
return _cache.rbegin()->second.get();
does not do what you want, as std::map
does not append elements but sorts them.不做你想做的,因为
std::map
不追加元素而是对它们进行排序。 However emplace
returns an iterator to what it just inserted, so you only need:但是
emplace
将迭代器返回到它刚刚插入的内容,因此您只需要:
return _cache.emplace(foo, std::make_unique<Bar>(bar))->first->second.get();
Or even, since you don't actually need to store and copy the Bar
, and you can also sacrifice foo
:甚至,由于您实际上并不需要存储和复制
Bar
,您也可以牺牲foo
:
return _cache.emplace(
std::move(foo),
std::make_unique<Bar>(ThirdPartLibrary::CreateBar())
)->first->second.get();
I'd also personally flip the (it == _cache.end())
condition to make it an early return, but that's just a matter of taste.我也会亲自翻转
(it == _cache.end())
条件以使其早日返回,但这只是品味问题。
Otherwise, what you have looks good to me.否则,你所拥有的对我来说看起来不错。
You tagged this as c++14, but for posterity I'll add a C++17 version:您将其标记为 c++14,但为了后代,我将添加一个 C++17 版本:
const Bar* TryGetBarValue(std::shared_ptr<Foo> foo)
{
struct DelayedBar
{
operator std::unique_ptr<Bar>() const { return std::make_unique<Bar>(thirdpartyLibrary::CreateBar()); }
};
return _cache.try_emplace(std::move(foo), DelayedBar()).first->second.get();
}
The try_emplace
function will emplace its arguments if the map doesn't already contain that key.如果映射尚未包含该键,则
try_emplace
函数将try_emplace
其参数。 If the key already exists, no object is constructed.如果键已经存在,则不构造任何对象。 In either case an iterator to that key/value pair is returned.
在任何一种情况下,都会返回该键/值对的迭代器。 This function avoids the double lookup involved when you do
find
-> emplace/insert
.当您确实
find
-> emplace/insert
时,此功能避免了涉及的双重查找。
In our case we can't simply pass the arguments of try_emplace
along so I've tried to be clever in delaying the construction of the object using this DelayedBar
class.在我们的例子中,我们不能简单地传递
try_emplace
的参数,所以我试图巧妙地使用这个DelayedBar
类延迟对象的构造。 It only calls CreateNewBar
when trying to cast to a std::unique_ptr<Bar>
which only happens when try_emplace
is trying to construct the object.它仅在尝试
try_emplace
为std::unique_ptr<Bar>
时调用CreateNewBar
,这仅在try_emplace
尝试构造对象时发生。
I have compiled this with GCC 8.2, Clang 7.0.0 and MSVC 19.16 (all via Compiler Explorer) and it compiles okay.我已经用 GCC 8.2、Clang 7.0.0 和 MSVC 19.16(全部通过编译器资源管理器)编译了它,它编译没问题。
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