C++ std::string 是否包含空终止符？
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Does std::string contain null terminator?
提问by mister
Will the below string contain the null terminator '\0' ?
下面的字符串会包含空终止符 '\0' 吗？
std::string temp = "hello whats up";
Thanks! :)
谢谢！:)
回答by jahhaj
No, but if you say temp.c_str()a null terminator will be included in the return from this method.
不，但是如果您说temp.c_str()空终止符将包含在此方法的返回中。
It's also worth saying that you can include a null character in a string just like any other character.
还值得一提的是，您可以像任何其他字符一样在字符串中包含空字符。
string s("hello");
cout << s.size() << ' ';
s[1] = 'std::string temp = "hello whats up";
';
cout << s.size() << '\n';
prints
印刷
5 5
5 5
and not 5 1as you might expect if null characters had a special meaning for strings.
5 1如果空字符对字符串具有特殊含义，则不会像您所期望的那样。
回答by mister
Not in C++03, and it's not even guaranteed before C++11 that in a C++ std::string is continuous in memory. Only C strings (char arrays which are intended for storing strings) had the null terminator.
不在 C++03 中，在 C++11 之前甚至不能保证 C++ std::string 在内存中是连续的。只有 C 字符串（用于存储字符串的字符数组）具有空终止符。
In C++11 and later, mystring.c_str()is equivalent to mystring.data()is equivalent to &mystring[0], and mystring[mystring.size()]is guaranteed to be '\0'.
在 C++11 及更高版本中，mystring.c_str()等效于mystring.data()等效于&mystring[0]，并且mystring[mystring.size()]保证为'\0'。
回答by aniliitb10
This depends on your definition of 'contain' here. In
这取决于您在此处对“包含”的定义。在
std::string s1 = "ab##代码####代码##cd"; // s1 contains "ab", using string literal
std::string s2{"ab##代码####代码##cd", 6}; // s2 contains "ab##代码####代码##cd", using different ctr
std::string s3 = "ab##代码####代码##cd"s; // s3 contains "ab##代码####代码##cd", using ""s operator
there are few things to note:
有几点需要注意：
temp.size()will return the number of characters from firsthto lastp(both inclusive)- But at the same time
temp.c_str()ortemp.data()will return with anullterminator - Or in other words
int(temp[temp.size()])will be zero
temp.size()将先返回字符数h去年p（均包括在内）- 但同时
temp.c_str()还是temp.data()会带着null终结符返回 - 或者换句话说
int(temp[temp.size()])将为零
I know, I sound similar to some of the answers here but I want to point out that sizeof std::stringin C++is maintained separately and it is not likein Cwhere you keep counting unless you find the first nullterminator.
我知道，我的声音有些类似于此的答案，但我想指出的是，size的std::string在C++被单独保存，它不是像在C哪里你保持，除非你找到的第一个计数null终止。
To add, the story would be a little different if your string literalcontains embedded \0. In this case, the construction of std::stringstops at first nullcharacter, as following:
另外，如果您string literal包含嵌入的\0. 在这种情况下，构造std::string在第一个null字符处停止，如下所示：
References:
参考：
回答by Nawaz
Yes if you call temp.c_str(), then it will return null-terminated c-string.
是的，如果您调用temp.c_str()，那么它将返回以空字符结尾的 c 字符串。
However, the actual data stored in the object tempmay not be null-terminated, but it doesn't matter and shouldn't matter to the programmer, because when then programmer wants const char*, he would call c_str()on the object, which is guaranteed to return null-terminated string.
但是，对象中存储的实际数据temp可能不是以null结尾的，但这对程序员来说无关紧要，因为当程序员想要时const char*，他会调用c_str()该对象，该对象保证返回null - 终止的字符串。
回答by Naps62
With C++ strings you don't have to worry about that, and it's possibly dependent of the implementation.
使用 C++ 字符串，您不必担心这一点，它可能取决于实现。
Using temp.c_str()you get a C representation of the string, which will definitely contain the \0char. Other than that, i don't really see how it would be useful on a C++ string
使用temp.c_str()您可以获得字符串的 C 表示，其中肯定会包含\0字符。除此之外，我真的不知道它在 C++ 字符串上有什么用
回答by Superman
std::stringinternally keeps a count of the number of characters. Internally it works using this count. Like others have said, when you need the string for display or whatever reason, you can its c_str()method which will give you the string with the null terminator at the end.
std::string内部保持字符数的计数。它在内部使用此计数工作。就像其他人所说的那样，当您需要显示字符串或出于任何原因时，您可以使用它的c_str()方法，该方法将为您提供末尾带有空终止符的字符串。
