C++ 风格从 unsigned char * 转换为 const char *

C++ style cast from unsigned char * to const char *

提问by Hymanhab

I have:

我有：

unsigned char *foo();
std::string str;
str.append(static_cast<const char*>(foo()));

The error: invalid static_cast from type ‘unsigned char*' to type ‘const char*'

错误： invalid static_cast from type ‘unsigned char*' to type ‘const char*'

What's the correct way to cast here in C++ style?

以 C++ 风格在此处进行转换的正确方法是什么？

采纳答案by Ruben Bartelink

reinterpret_cast

reinterpret_cast

回答by Brian R. Bondy

char *and const unsigned char *are considered unrelated types. So you want to use reinterpret_cast.

char *并且const unsigned char *被认为是不相关的类型。所以你想使用reinterpret_cast.

But if you were going from const unsigned char*to a non consttype you'd need to use const_castfirst. reinterpret_castcannot cast away a constor volatilequalification.

但是，如果您要从const unsigned char*非const类型转换，则需要先使用const_cast。reinterpret_cast不能抛弃 aconst或volatile资格。

回答by JaredPar

Try reinterpret_cast

尝试 reinterpret_cast

unsigned char *foo();
std::string str;
str.append(reinterpret_cast<const char*>(foo()));

回答by AJG

unsigned char* is basically a byte array and should be used to represent raw data rather than a string generally. A unicode string would be represented as wchar_t*

unsigned char* 基本上是一个字节数组，通常应该用于表示原始数据而不是字符串。Unicode 字符串将表示为 wchar_t*

According to the C++ standard a reinterpret_cast between unsigned char* and char* is safe as they are the same size and have the same construction and constraints. I try to avoid reintrepret_cast even more so than const_cast in general.

根据 C++ 标准， unsigned char* 和 char* 之间的 reinterpret_cast 是安全的，因为它们具有相同的大小并具有相同的构造和约束。我尽量避免 reintrepret_cast 比一般的 const_cast 更是如此。

If static cast fails with what you are doing you may want to reconsider your design because frankly if you are using C++ you may want to take advantage of what the "plus plus" part offers and use string classes and STL (aka std::basic_string might work better for you)

如果静态转换因您正在执行的操作失败，您可能需要重新考虑您的设计，因为坦率地说，如果您使用 C++，您可能希望利用“加号”部分提供的功能并使用字符串类和 STL（又名 std::basic_string可能更适合你）

回答by Timo Geusch

You would need to use a reinterpret_cast<>as the two types you are casting between are unrelated to each other.

您需要使用 a ，reinterpret_cast<>因为您在两者之间进行转换的两种类型彼此无关。

回答by Victor Sergienko

Too many comments to make to different answers, so I'll leave another answer here.

太多的评论无法对不同的答案进行评论，所以我会在这里留下另一个答案。

You can and should use reinterpret_cast<>, in your case

你可以而且应该使用reinterpret_cast<>，在你的情况下

str.append(reinterpret_cast<const char*>(foo()));

because, while these two are different types, the 2014 standard, chapter 3.9.1 Fundamental types [basic.fundamental]says there isa relationship between them:

因为，虽然这两个是不同的类型，但 2014 标准，章节3.9.1 Fundamental types [basic.fundamental]说它们之间存在关系：

Plain char, signed charand unsigned charare three distinct types, collectively called narrow character types. A char, a signed char, and an unsigned charoccupy the same amount of storage and have the same alignment requirements (3.11); that is, they have the same object representation.

平原char，signed char并unsigned char有三种不同的类型，统称为窄字符类型。A char、 asigned char和 anunsigned char占用相同的存储量并具有相同的对齐要求（3.11）；也就是说，它们具有相同的对象表示。

(selection is mine)

（选择是我的）

Here's an available link: https://en.cppreference.com/w/cpp/language/types#Character_types

这是一个可用的链接：https: //en.cppreference.com/w/cpp/language/types#Character_types

Using wchar_tfor Unicode/multibyte strings is outdated: Should I use wchar_t when using UTF-8?

使用wchar_t：对Unicode /多字节字符串已经过时使用UTF-8的时候我应该使用wchar_t的？

回答by joi

Hope it help. :)

希望有帮助。:)

const unsigned attribName = getname();
const unsigned attribVal = getvalue();
const char *attrName=NULL, *attrVal=NULL;
attrName = (const char*) attribName;
attrVal = (const char*) attribVal;