字符串是否包含空终止符?

58  

No, but if you say temp.c_str() a null terminator will be included in the return from this method.

没有，但是如果您说temp.c_str()，那么这个方法的返回中将包含一个空终止符。

It's also worth saying that you can include a null character in a string just like any other character.

同样值得说的是，您可以在字符串中包含一个空字符，就像其他字符一样。

string s("hello");
cout << s.size() << ' ';
s[1] = '\0';
cout << s.size() << '\n';

prints

打印

5 5

5个5

and not 5 1 as you might expect if null characters had a special meaning for strings.

如果空字符对于字符串有特殊的含义，就不像你想象的那样是51。

33  

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.

在c++ 03中没有，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null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中，mystr.c str()等价于mystring.data()等价于&mystring[0]， mystring[mystr.size()]保证为'\0'。

2  

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null结尾的c-string。

However, the actual data stored in the object temp may 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temp中存储的实际数据可能不是以null结尾的，但这对程序员来说并不重要，也不应该重要，因为当程序员需要const char*时，他会在对象上调用c_str()，该对象保证返回以null结尾的字符串。

2  

This depends on your definition of 'contain' here. In

这取决于你对“包含”的定义。在

std::string temp = "hello whats up";

there are few things to note:

有几件事需要注意:

• temp.size() will return the number of characters from first h to last p
• size()将返回从第一个h到最后一个p的字符数
• But at the same time temp.c_str() or temp.data() will return with a null terminator
• 但与此同时，temp.c_str()或temp.data()将返回一个空终止符
• Or in other words int(temp[temp.size()]) will be zero
• 换句话说，int(temp .size())将为零。

I know, I sound similar to some of the answers here but I want to point out that size of std::string in C++ is maintained separately and it is not like in C where you keep counting unless you find the first null terminator.

我知道，我听起来和这里的一些答案相似，但我想指出的是，在c++中，std::string的大小是分开维护的，不像在C中，你一直计数，除非你找到第一个空终止符。

To add, the story would be a little different if your string literal contains embedded \0. In this case, the construction of std::string stops at first null character, as following:

若要添加，如果字符串文字包含嵌入的\0，则情况会有所不同。在这种情况下，std::string的构造在第一个空字符处停止，如下所示:

std::string s1 = "ab\0\0cd";   // s1 contains "ab",       using string literal
std::string s2{"ab\0\0cd", 6}; // s2 contains "ab\0\0cd", using different ctr
std::string s3 = "ab\0\0cd"s;  // s3 contains "ab\0\0cd", using ""s operator

References:

引用:

• https://akrzemi1.wordpress.com/2014/03/20/strings-length/
• https://akrzemi1.wordpress.com/2014/03/20/strings-length/
• https://en.cppreference.com/w/cpp/string/basic_string/basic_string
• https://en.cppreference.com/w/cpp/string/basic_string/basic_string

1  

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，您不必担心这个问题，它可能依赖于实现。

Using temp.c_str() you get a C representation of the string, which will definitely contain the \0 char. Other than that, i don't really see how it would be useful on a C++ string

使用temp.c_str()，可以得到字符串的C表示形式，它肯定包含\0 char。除此之外，我不知道它在c++字符串中有什么用

1  

std::string internally 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.

字符串内部保存字符数。在内部，它使用这个计数。就像其他人说的，当你需要字符串显示或其他原因时，你可以使用它的c_str()方法，它会给你一个字符串，在末尾有空终止符。