转换（无效*）到std ::矢量＆lt; unsigned char型＆GT; [英] Converting (void*) to std::vector<unsigned char>

问题描述

我有一个（无效*）缓冲，我需要转换为的std ::矢量＆lt; unsigned char型＆GT; 之前，我可以通过它。不幸的是，我的C ++的铸造技能稍微弱。有什么建议？

I have a (void*) buffer that I need to convert to std::vector<unsigned char> before I can pass it on. Unfortunately, my C++ casting skills a little weak. Any suggestions?

推荐答案

您需要的缓冲区的长度。一旦你这样做，我们可以做到这一点：

You will need the length of the buffer. Once you do, we can do this:

unsigned char *charBuf = (unsigned char*)voidBuf;
/* create a vector by copying out the contents of charBuf */
std::vector<unsigned char> v(charBuf, charBuf + len);

好吧，评论让我开始对我为什么不使用 reinter pret_cast ：

  

• 在C ++中，C样式转换是一个方便的功能 - 它要求编译器选择转换的最安全，最便携的形式在一组可用的转换操作符的

• In C++, the C-style cast is a convenience function -- it asks the compiler to choose the safest and most portable form of conversion over the set of available cast operators.

的 reinter pret_cast 是实现定义，应该永远是你心中的最后一件事（和使用时，你一定做一个非便携式事情明知）。

The reinterpret_cast is implementation defined and should always be the last thing on your mind (and used when you are necessarily doing a non-portable thing knowingly).

（无符号不改变类型）的char * 和<$ C之间的转换$ C>无效* 是可移植的（你可以实际使用的static_cast 如果你真的挑剔）。

The conversion between (unsigned doesn't change the type) char * and void * is portable (you could actually use static_cast if you are really picky).

与C样式转换的问题是：增加了灵活性可能会导致心痛，当指针类型更改

The problem with the C-style cast is: the added flexibility can cause heartaches when the pointer type changes.

注意：我同意不铸造尽可能多的一般惯例。然而，如果没有提供的任何来源，这是我能做的最好的。

Note: I agree with the general convention of not casting as much as possible. However, without any source provided, this is the best I could do.

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