是否可以假定STL向量存储始终是连续的？ [英] Is it safe to assume that STL vector storage is always contiguous?

问题描述

如果你有一个已经调整大小的STL向量，是否可以取元素0的地址，并假设向量的其余部分将在内存中？

If you have an STL vector which has been resized, is it safe to take the address of element 0 and assume the rest of the vector will follow in memory?

例如

vector<char> vc(100);
// do some stuff with vc
vc.resize(200);
char* p = &vc[0];
// do stuff with *p

推荐答案

从C ++ 03标准（23.2.4.1）：

From the C++03 standard (23.2.4.1):

向量的元素连续存储
，如果v是
向量，其中T是除了bool之外的某些
类型，则它遵守
身份& v [n] ==& v [0] + n for
all 0< = n<

The elements of a vector are stored contiguously, meaning that if v is a vector where T is some type other than bool, then it obeys the identity &v[n] == &v[0] + n for all 0 <= n < v.size().

但是请注意，在向其添加元素之后，数组将被重新分配（无效任何指针和迭代器）。

But watch out for the array being reallocated (invalidating any pointers and iterators) after adding elements to it.

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