问题描述
我在一个嵌入式系统的某些C ++中包装了用户空间的linux socket功能（是的，这可能是重新发明轮子）。
I'm wrapping up user space linux socket functionality in some C++ for an embedded system (yes, this is probably reinventing the wheel again).
使用向量提供读写实现。
I want to offer a read and write implementation using a vector.
写操作很容易，我可以通过& myvec [0] code>并避免不必要的复制。我想做同样的事情，直接读入一个向量，而不是读入一个字符缓冲区，然后复制到一个新创建的向量。
Doing the write is pretty easy, I can just pass &myvec[0] and avoid unnecessary copying. I'd like to do the same and read directly into a vector, rather than reading into a char buffer then copying all that into a newly created vector.
现在，我知道我想要读多少数据，我可以适当地分配（ vec.reserve（））。我也可以读入& myvec [0] ，虽然这可能是一个非常糟糕的想法。显然这样做不允许myvec.size返回任何合理的。有什么办法这样做：
Now, I know how much data I want to read, and I can allocate appropriately (vec.reserve()). I can also read into &myvec[0], though this is probably a VERY BAD IDEA. Obviously doing this doesn't allow myvec.size to return anything sensible. Is there any way of doing this that:
- 从安全/ C ++的角度来看，并不完全感觉到幸运
- 不涉及数据块的两个副本 - 一次从内核到用户空间，一次从C
char *样式缓冲区到C ++向量。 li>
- Doesn't completely feel yucky from a safety/C++ perspective
- Doesn't involve two copies of the data block - once from kernel to user space and once from a C
char *style buffer into a C++ vector.
推荐答案
使用 resize（） reserve（）。这将正确设置向量的大小 - 之后，& myvec [0] 像往常一样，保证指向一个连续的内存块。
Use resize() instead of reserve(). This will set the vector's size correctly -- and after that, &myvec[0] is, as usual, guaranteed to point to a continguous block of memory.
编辑：使用& myvec [0] 作为指向读写基础数组的指针是安全和有保证的以C ++标准工作。这里是Herb Sutter的： / p>
Using &myvec[0] as a pointer to the underlying array for both reading and writing is safe and guaranteed to work by the C++ standard. Here's what Herb Sutter has to say:
这篇关于将read（）直接使用到C ++ std：vector中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！