By taking a C string the C++03 std::fstream class reduced dependency on the std::string class. In C++11, however, the std::fstream class does allow passing a std::string for its constructor parameter. Now, you may wonder why isn’t there a transparent conversion from a std:string to a C string, so a class that expects a C string … Read more
The problem is that a literal 0 is a null pointer constant. The compiler doesn’t know if you meant: std::string::operator +=(const char*); // tmp += “abc”; or std::string::operator +=(char); // tmp += ‘a’; (better compilers list the options). The workround (as you have discovered) is to write the append as: tmp += ‘\0’; (I assume … Read more
std::wstring ws( args.OptionArg() ); std::string test( ws.begin(), ws.end() );
A std::string’s allocation is not guaranteed to be contiguous under the C++98/03 standard, but C++11 forces it to be. In practice, neither I nor Herb Sutter know of an implementation that does not use contiguous storage. Notice that the &s[0] thing is always guaranteed to work by the C++11 standard, even in the 0-length string … Read more
Unfortunately, this is UB, if I interpret the wording correct (in any case, it’s not allowed): §21.4.5 [string.access] p2 Returns: *(begin() + pos) if pos < size(), otherwise a reference to an object of type T with value charT(); the referenced value shall not be modified. (Editorial error that it says T not charT.) .data() … Read more
TL;DR operator+= is a class member function in class string, while operator+ is a template function. The standard class template<typename CharT> basic_string<CharT> has overloaded function basic_string& operator+=(CharT), and string is just basic_string<char>. As values that fits in a lower type can be automatically cast into that type, in expression s += 2, the 2 is … Read more
This is an intentional decision in libc++’s implementation of std::string. First of all, std::string has so-called Small String Optimization (SSO), which means that for very short (or empty) strings, it will store their contents directly inside of the container, rather than allocating dynamic memory. That’s why we don’t see any allocations in either case. In … Read more
Well, you could erase() the first character too (note that erase() modifies the string): m_VirtualHostName.erase(0, 1); m_VirtualHostName.erase(m_VirtualHostName.size() – 1); But in this case, a simpler way is to take a substring: m_VirtualHostName = m_VirtualHostName.substr(1, m_VirtualHostName.size() – 2); Be careful to validate that the string actually has at least two characters in it first…
Consider this: std::string str = “Hello ” + “world”; // bad! Both the rhs and the lhs for operator + are char*s. There is no definition of operator + that takes two char*s (in fact, the language doesn’t permit you to write one). As a result, on my compiler this produces a “cannot add two … Read more
Converting to a std::string requires a “user defined conversion”. Converting to void const* does not. User defined conversions are ordered behind built in ones.