c++ - 如何初始化STL容器中存在的类的数据？

Question

例如，我有 A 类:

 class A{
           int value_;
           public:
           A(A& a){
                value_ = a.value_;
           }
           A(int value){
                value_ = value;
           }
 };

我想要一个 A 类 vector ，但我想将所有 vector 的值传递给 A(int value)。

 std::vector<A,allocator<A>> my_vector;

• 最好的方法是什么？
• 有没有办法使用分配器？

Answer 1

c++11

With the new standard added functionality was granted to objects of Allocator type.

One of the features added was that Allocators now allows emplacement construction, aka. construction of objects using a constructor other than copy/move.

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template< class U, class... Args > void construct( U* p, Args&&... args );

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The standard does guarantee that STL containers must use this new feature, and with that said you could implement your own allocator just for the purpose of default initializing a non-default-initializable object.

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^{It's not the prettiest solution, but whatever floats your boat..}

c++03

The allocator has nothing to do with that part of object initialization, it's only purpose is to allocate/deallocate memory, the type of initialization you are referring to is done elsewhere.

The only constructor an allocator will call is the copy-constructor when someone asks it to perform a placement-new, and the value passed to that copy-ctor has already been established somewhere else.

To sum things up; No, you cannot use an allocator so solve this particular problem.

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std::vector 什么时候需要使用default-ctor？

std::vector仅在两种情况下使用它拥有的类型的默认构造函数:

1. 您指定 std::vector 的元素数量在适当的构造函数重载中，但不提供默认值

2. 您使用 std::vector<T>::resize (n)并增加容器中对象的数量(注意没有指定成员函数的第二个参数)

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考虑到上述内容，我们可以使用容器做很多事情，而无需在我们的对象中提供默认构造函数，例如将其初始化为包含 N 个值为 X 的元素.

struct A{
  A (A const& a)
   : value_ (a.value_) 
  { } 

  A (int value)
    : value_ (value)
  {}  

  int value_;
};

int
main (int argc, char *argv[])
{
  std::vector<A> vec (5, A(1)); // initialize vector with 5 elements of A(1)

  vec.push_back (A(3));         // add another element
}

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但是我真的很想能够使用vec.resize()!?

然后你有二、三、四个选项:

1. 使用分配器的 C++11 方法

2. 让你的对象有一个默认构造函数

3. 用一个非常薄的包装器包装您的对象，其唯一目的是默认初始化包含的对象(这在某些情况下说起来容易做起来难)

4. “在 boost::optional 中包装 [对象] 实际上可以为任何类型提供默认构造函数” - @ Xeo