是否& s [0]指向std :: string中的连续字符？ [英] Does "&s[0]" point to contiguous characters in a std::string?

问题描述

我正在进行一些维护工作，并且遇到以下情况：

I'm doing some maintenance work and ran across something like the following:

std::string s;
s.resize( strLength );  
// strLength is a size_t with the length of a C string in it. 

memcpy( &s[0], str, strLength );

我知道使用& s [0]是安全的，如果它是一个std :: vector，但是这是一个安全使用std :: string？

I know using &s[0] would be safe if it was a std::vector, but is this a safe use of std::string?

推荐答案

std :: string的分配不保证是连续的C ++ 98/03标准，但C ++ 11强制它。实际上，我和 Herb Sutter 知道不使用连续存储的实现。

A std::string's allocation is not guaranteed to be contiguous under the C++98/03 standard, but C++11 forces it to be. In practice, neither I nor Herb Sutter know of an implementation that does not use contiguous storage.

请注意，& s [0] 总是保证通过C ++ 11标准工作，即使在0长度字符串的情况下。如果您 str.begin（）或& * str.begin（），但不能保证& s [0] 标准定义 operator [] 为：

Notice that the &s[0] thing is always guaranteed to work by the C++11 standard, even in the 0-length string case. It would not be guaranteed if you did str.begin() or &*str.begin(), but for &s[0] the standard defines operator[] as:

返回： *（begin（）+ pos） if pos < size（），否则引用 T ，值为 charT（）;参考值不应修改

Returns: *(begin() + pos) if pos < size(), otherwise a reference to an object of type T with value charT(); the referenced value shall not be modified

继续， data（）定义为：

返回：指针 p 使中的每个 i 的 p + i ==& operator []（i） > [0，size（）] 。

Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].

（注意范围两端的方括号）

(notice the square brackets at both ends of the range)

注意：预标准化C ++ 0x不能保证& s [0] 使用零长度字符串（实际上，它是明确未定义的行为），而这个答案的旧版本解释了这一点;这在以后的标准草案中已经解决，因此答案已相应更新。

Notice: pre-standardization C++0x did not guarantee &s[0] to work with zero-length strings (actually, it was explicitly undefined behavior), and an older revision of this answer explained this; this has been fixed in later standard drafts, so the answer has been updated accordingly.

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