std :: vector of std :: vectors contiguity [英] std::vector of std::vectors contiguity

问题描述

我知道 std :: vector< T> 在内部连续地存储它的数据（除非 std :: vector< bool> code>标准和 C ++ 11 标准中的 / p>



很好的stackoverflow问题，处理这个问题并引用标准： answer ，回答。

嵌套向量中的数据 std :: vector< std :: vector< T> > ？如何存储？

如果每个内部向量都需要连续存储数据，那么& v [n] = =& v [0] + n，对于所有0 <= n < v.size（）。

要对此略有不同，可以访问所有元素在这样的嵌套结构中简单地并且顺序地（通过指针或类似物）以与对于1-D向量可以完成的相同的方式进行？

解决方案

否。 向量的元素存储在动态分配的存储器块中;否则，向量的容量不能增加。 向量对象只保存指向该块的指针。

元素按顺序存储的要求仅适用于元素本身，而不适用于这些元素的任何动态分配成员。

I know that std::vector<T> internally stores it's data contiguously (unless it is std::vector<bool>) both in the old C++03 standard and the new C++11.

Nice stackoverflow questions that deal with this and quote the standard: answer, answer.

What about the data inside nested vectors std::vector <std::vector <T> >? How is that stored?

If every internal vector needs to store it's data contiguously, how can it be true that &v[n] == &v[0] + n for all 0 <= n < v.size().

To phrase this slightly different, is it possible to access all the elements stored in such nested structure "simply" and sequentially (via a pointer or similar) the same way it can be done for a 1-D vector?

解决方案

No. The elements of a vector are stored in a dynamically allocated block of memory; otherwise, the capacity of the vector could not increase. The vector object just holds a pointer to that block.

The requirement that the elements be stored sequentially applies only to the elements themselves, and not to any dynamically allocated members of those elements.

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