问题描述
如果您有一个已调整大小的 STL 向量，那么获取元素 0 的地址并假设向量的其余部分将跟随在内存中是否安全?
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
是的，这是一个有效的假设 (*).
来自 C++03 标准 (23.2.4.1):
From the C++03 standard (23.2.4.1):
存储向量的元素连续，这意味着如果 v 是向量，其中 T 是一些类型不是 bool，那么它服从身份 &v[n] == &v[0] + n for所有 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().
(*) ...但要注意在向数组添加元素后重新分配数组(使任何指针和迭代器无效).
(*) ... but watch out for the array being reallocated (invalidating any pointers and iterators) after adding elements to it.
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