How to resize std::string to remove all null terminator characters?

Question

I'm using a std::string to interface with a C-library that requires a char* and a length field:

std::string buffer(MAX_BUFFER_SIZE, '\0');
TheCLibraryFunction(&buffer[0], buffer.size());

However, the size() of the string is the actual size, not the size of the string containing actual valid non-null characters (i.e. the equivalent of strlen()). What is the best method of telling the std::string to reduce its size so that there's only 1 ending null terminator character no explicit null terminator characters? The best solution I can think of is something like:

buffer.resize(strlen(buffer.c_str()));

Or even:

char buffer[MAX_BUFFER_SIZE]{};
TheCLibraryFunction(buffer, sizeof(buffer));
std::string thevalue = buffer;

Hoping for some built-in / "modern C++" way of doing this.

EDIT

I'd like to clarify the "ending null terminator" requirement I mentioned previously. I didn't mean that I want 1 null terminator explicitly in std::string's buffer, I was more or less thinking of the string as it comes out of basic_string::c_str() which has 1 null terminator. However for the purposes of the resize(), I want it to represent the size of actual non-null characters. Sorry for the confusion.

Answer 1

Many ways to do this; but probably the one to me that seems to be most "C++" rather than C is:

str.erase(std::find(str.begin(), str.end(), '\0'), str.end());

i.e. Erase everything from the first null to the end.

Answer 2

You can do this:

buffer.erase(std::find(buffer.begin(), buffer.end(), '\0'), buffer.end());

Consider std::basic_string::erase has an overloading:

basic_string& erase( size_type index = 0, size_type count = npos );

A more succinct way:

buffer.erase(buffer.find('\0'));