blocks|key|660376|text|std::string对字节进行操作，而不是对Unicode字符进行操作，因此std::string::size()确实会以字节为单位返回数据的大小(当然，没有std::string存储数据所需的开销)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|660377|不需要，std::string只存储您告诉它存储的数据(它不需要尾随的NULL字符)。因此，它不会包含在大小中，除非您显式创建了一个尾随NULL字符的字符串。|660378|entityMap^0|0|B|13|J|28|B|0|4|B|Z|4|1W|4|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|I|$]]

<code>std::string</code> operates on bytes, not on Unicode characters, so <code>std::string::size()</code> will indeed return the size of the data in bytes (without the overhead that <code>std::string</code> needs to store the data, of course).

No, <code>std::string</code> stores only the data you tell it to store (it does not need the trailing <code>NULL</code> character). So it will not be included in the size, unless you explicitly create a string with a trailing <code>NULL</code> character.

blocks|key|877667|text|你可能会对此有点学究：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|877668|std::string+x("X");

std::cout+<<+x.size()+*+sizeof(std::string::value_type);|code-block|syntax|javascript|877669|但是std::string::value_type是字符，并且sizeof(字符)被定义为1。|877670|这只在你定义字符串类型时才变得重要(因为它在将来可能会改变，或者因为编译器选项)。|877671|//+Some+header+file:
typedef+++std::basic_string<T_CHAR>++T_string;

//+Source+a+million+miles+away
T_string+++x("X");

std::cout+<<+x.size()+*+sizeof(T_string::value_type);|877672|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You could be pedantic about it:

<pre><code>std::string x("X");

std::cout &lt;&lt; x.size() * sizeof(std::string::value_type);
</code></pre>

But std::string::value_type is char and sizeof(char) is defined as 1.

This only becomes important if you typedef the string type (because it may change in the future or because of compiler options).

<pre><code>// Some header file:
typedef std::basic_string&lt;T_CHAR&gt; T_string;

// Source a million miles away
T_string x("X");

std::cout &lt;&lt; x.size() * sizeof(T_string::value_type);
</code></pre>

blocks|key|660443|text|std::string::size()实际上是以字节为单位的大小。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|660444|entityMap^0|0|J|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>std::string::size()</code> is indeed the size in bytes.

blocks|key|660510|text|要获得字符串使用的内存量，必须将capacity()与用于管理的开销相加。请注意，它是capacity()而不是size()。容量决定了分配的字符数(charT)，而size()则告诉您实际使用了多少个字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|660511|特别是，std::string实现通常不会*shrink_to_fit*内容，所以如果您创建一个字符串，然后从末尾删除元素，size()将会递减，但在大多数情况下(这是由实现定义的)+capacity()不会。|660512|某些实现可能不会分配所需的确切内存量，而是获取给定大小的块以减少内存碎片。在将两个大小的块的功率用于字符串的实现中，大小为17的字符串可以分配与32字符一样多的字符。|660513|entityMap^0|G|A|17|A|1K|6|23|5|2B|6|0|4|B|1Q|6|2K|A|0|1P|2|20|2|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|K|$]]

To get the amount of memory in use by the string you would have to sum the <code>capacity()</code> with the overhead used for management. Note that it is <code>capacity()</code> and not <code>size()</code>. The capacity determines the number of characters (<code>charT</code>) allocated, while <code>size()</code> tells you how many of them are actually in use.

In particular, <code>std::string</code> implementations don't usually *shrink_to_fit* the contents, so if you create a string and then remove elements from the end, the <code>size()</code> will be decremented, but in most cases (this is implementation defined) <code>capacity()</code> will not. 

Some implementations might not allocate the exact amount of memory required, but rather obtain blocks of given sizes to reduce memory fragmentation. In an implementation that used power of two sized blocks for the strings, a string with size <code>17</code> could be allocating as much as <code>32</code> characters.

blocks|key|660421|text|是的，char+()会给出字符串中的大小。多字节编码中的一个字符占用多个char。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|660422|entityMap^0|3|4|10|4|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Yes, size() will give you the number of <code>char</code> in the string. One character in multibyte encoding take up multiple <code>char</code>.

blocks|key|660573|text|问题本身就存在冲突：std::string被定义为std::basic_string<char,...>+--也就是说，它的元素类型是char+(1字节)，但是后来您声明“字符串包含多字节字符串”("+multibyte“==+wchar_t?)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|660574|size()成员函数不计算尾随null。它的值表示字符数(而不是字节数)。|660575|假设您打算将多字节字符串指定为std::wstring+(std::basic_string<wchar_t,...>的别名)，则std::wstring字符(包括空终止符)的内存占用空间为：|660576|std::wstring+myString;
+...
size_t+bytesCount+=+(myString.size()+%2B+1)+*+sizeof(wchar_t);|code-block|syntax|javascript|660577|考虑如何编写一个可重用的模板函数来处理任何可能的std::basic_string<>实例化都是很有指导意义的**|660578|//+Return+number+of+bytes+occupied+by+null-terminated+inString.c_str().
template+<typename+_Elem>
inline+size_t+stringBytes(const+std::basic_string<typename+_Elem>&+inString,+bool+bCountNull)
{
+++return+(inString.size()+%2B+(bCountNull+?+1+:+0))+*+sizeof(_Elem);
}|660579|**为简单起见，忽略很少为std::basic_string<>显式指定的特征和分配器类型(它们具有默认值)。|660580|entityMap^0|A|B|P|R|1V|4|36|7|0|0|6|0|F|C|T|U|1T|C|0|0|0|0|D|J|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|1F|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1G|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1H|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|V|$]]

There is inherent conflict in the question as written: <code>std::string</code> is defined as <code>std::basic_string&lt;char,...&gt;</code> -- that is, its element type is <code>char</code> (1-byte), but later you stated "the string contains a multibyte string" ("multibyte" == <code>wchar_t</code>?).

The <code>size()</code> member function does not count a trailing null. It's value represents the number of characters (not bytes).

Assuming you intended to say your multibyte string is <code>std::wstring</code> (alias for <code>std::basic_string&lt;wchar_t,...&gt;</code>), the memory footprint for the <code>std::wstring</code>'s characters, including the null-terminator is:

<pre><code>std::wstring myString;
 ...
size_t bytesCount = (myString.size() + 1) * sizeof(wchar_t);
</code></pre>

It's instructive to consider how one would write a reusable template function that would work for ANY potential instantiation of std::basic_string&lt;> like this**:

<pre><code>// Return number of bytes occupied by null-terminated inString.c_str().
template &lt;typename _Elem&gt;
inline size_t stringBytes(const std::basic_string&lt;typename _Elem&gt;&amp; inString, bool bCountNull)
{
 return (inString.size() + (bCountNull ? 1 : 0)) * sizeof(_Elem);
}
</code></pre>

** For simplicity, ignores the traits and allocator types rarely specified explicitly for <code>std::basic_string&lt;&gt;</code> (they have defaults).

I would like to get the bytes a <code>std::string</code>'s string occupies in memory, not the number of characters. The string contains a multibyte string. Would <code>std::string::size()</code> do this for me?

EDIT: Also, does <code>size()</code> also include the terminating <code>NULL</code>?

Get size of a std::string's string in bytes

编译器

我想要得到一个std::string的字符串在内存中所占的字节数，而不是字符数。这个字符串包含一个多字节的字符串。std::string::size()能帮我做到这一点吗？编辑:另外，size()是否也包括终止NULL

问获取std：：string字符串的字节大小
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