blocks|key|1327098|text|vector总是复制存储在向量中的任何内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327099|如果您保留了一个指针向量，那么它将复制指针，但不会复制指针所指向的实例。如果你正在处理大对象，你可以(也许应该)总是使用一个指针向量。通常，出于安全目的，使用适当类型的智能指针向量是很好的，否则处理对象生存期和内存管理可能会很棘手。|1327100|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

std::vector always makes a copy of whatever is being stored in the vector.

If you are keeping a vector of pointers, then it will make a copy of the pointer, but not the instance being to which the pointer is pointing. If you are dealing with large objects, you can (and probably should) always use a vector of pointers. Often, using a vector of smart pointers of an appropriate type is good for safety purposes, since handling object lifetime and memory management can be tricky otherwise.

blocks|key|1111930|text|是的，std::vector<T>::push_back()会创建参数的副本并将其存储在向量中。如果要在矢量中存储指向对象的指针，请创建一个std::vector<whatever*>而不是std::vector<whatever>。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1111931|但是，您需要确保指针引用的对象仍然有效，而向量持有对它们的引用(智能指针利用RAII习惯用法解决了这个问题)。|1111932|entityMap^0|3|R|1Y|M|2N|L|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

Yes, <code>std::vector&lt;T&gt;::push_back()</code> creates a copy of the argument and stores it in the vector. If you want to store pointers to objects in your vector, create a <code>std::vector&lt;whatever*&gt;</code> instead of <code>std::vector&lt;whatever&gt;</code>. 

However, you need to make sure that the objects referenced by the pointers remain valid while the vector holds a reference to them (smart pointers utilizing the RAII idiom solve the problem).

blocks|key|1111993|text|不仅std::vector会对您所推回的内容进行复制，而且集合的定义也会声明它会这样做，而且如果向量中没有正确的复制语义，就不能使用对象。因此，例如，您不能在向量中使用auto_ptr。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1111994|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Not only does std::vector make a copy of whatever you're pushing back, but the definition of the collection states that it will do so, and that you may not use objects without the correct copy semantics within a vector. So, for example, you do not use auto_ptr in a vector.

blocks|key|1327226|text|如果您不想要副本，那么最好的方法是使用指针向量(或用于相同目标的另一个结构)。如果需要副本，可以直接使用push_back()。你别无选择。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327227|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

if you want not the copies; then the best way is to use a pointer vector(or another structure that serves for the same goal).
if you want the copies; use directly push_back().
you dont have any other choice.

blocks|key|1327268|text|为什么要进行大量的valgrind调查才能找出答案！你可以用一些简单的代码来证明这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1327269|std::vector<std::string>+vec;

{
++++++std::string+obj("hello+world");
++++++vec.push_pack(obj);
}

std::cout+<<+vec[0]+<<+std::endl;++|code-block|syntax|javascript|1327270|如果打印了"hello+world“，则该对象必须已被复制|1327271|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Why did it take a lot of valgrind investigation to find this out!
Just prove it to yourself with some simple code e.g.

<pre><code>std::vector&lt;std::string&gt; vec;

{
 std::string obj("hello world");
 vec.push_pack(obj);
}

std::cout &lt;&lt; vec[0] &lt;&lt; std::endl; 
</code></pre>

If "hello world" is printed, the object must have been copied

blocks|key|1327328|text|与C%2B%2B11相关的是emplace成员函数系列，它允许您通过将对象移动到容器中来转移对象的所有权。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1327329|这个习惯用法看起来像这样|1327330|std::vector<Object>+objs;

Object+l_value_obj+{+/*+initialize+*/+};
//+use+object+here...

objs.emplace_back(std::move(l_value_obj));|code-block|syntax|javascript|1327331|左值对象的移动很重要，否则它将作为引用或常量引用转发，并且不会调用移动构造函数。|1327332|entityMap^0|A|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

Relevant in C++11 is the <code>emplace</code> family of member functions, which allow you to transfer ownership of objects by moving them into containers.

The idiom of usage would look like

<pre><code>std::vector&lt;Object&gt; objs;

Object l_value_obj { /* initialize */ };
// use object here...

objs.emplace_back(std::move(l_value_obj));
</code></pre>

The move for the lvalue object is important as otherwise it would be forwarded as a reference or const reference and the move constructor would not be called.

After a lot of investigations with valgrind, I've made the conclusion that std::vector makes a copy of an object you want to push_back.

Is that really true ? A vector cannot keep a reference or a pointer of an object without a copy ?!

Thanks

Is std::vector copying the objects with a push_back?

内存管理

容器

在对valgrind进行了大量研究之后，我得出结论: std::vector创建了您想要push_back的对象的副本。这是真的吗？如果没有副本，向量就不能保留对象的引用或指针?！谢谢

问是std::vector使用push_back复制对象吗？
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