In C++ strings are copied until a NULL character is received while feeding in a sequence of characters. But if you supply the number of characters to be read, will it copy past the NULL character? I have a situation where I may receive a message that has a NULL character in the middle and still useful information after it. The same question applies to append.
Similarly for find(), will it stop searching if it hits a NULL character?
You should be able to construct a string containing the '\0' character:
const char a[] = "Hello\0world";
std::string s(a, sizeof(a));
std::cout << "a = \"" << a << "\"\n";
std::cout << "s = \"" << s << "\"\n";
std::cout << "sizeof(a) = " << sizeof(a) << '\n';
std::cout << "strlen(a) = " << std::strlen(a) << '\n';
std::cout << "s.length() = " << s.length() << '\n';
The above snippet will print
a = "Hello" s = "Helloworld" sizeof(a) = 12 strlen(a) = 5 s.length() = 12
std::string doesn't know \0 as a string terminator! Poor C-style strings :-Dstd::basic_streambuf::sputn with the whole string, including the embedded '\0'. My guess is that the '\0' is dropped after the library have written the string to the operating system.std::string doesn't care about the C-style terminator, it has its length instead. If you don't provide a length on construction, then of course the string will infer it from finding the C-style terminator character.
std::stringthen yes, they can have \0 in the middle. std::string don't use the "ended by \0" convention. They have memory for storing the size. So they can be used as buffers.