Reading/writing/printing UTF-8 in C++11

Ephemera picture Ephemera · Mar 18, 2013 · Viewed 8.7k times · Source

I have been exploring C++11's new Unicode functionality, and while other C++11 encoding questions have been very helpful, I have a question about the following code snippet from cppreference. The code writes and then immediately reads a text file saved with UTF-8 encoding.

// Write
std::ofstream("text.txt") << u8"z\u6c34\U0001d10b";

// Read
std::wifstream file1("text.txt");
file1.imbue(std::locale("en_US.UTF8"));
std::cout << "Normal read from file (using default UTF-8/UTF-32 codecvt)\n";
for(wchar_t c; file1 >> c; ) // ?
   std::cout << std::hex << std::showbase << c << '\n';

My question is quite simply, why is a wchar_t needed in the for loop? A u8 string literal can be declared using a simple char * and the bit layout of the UTF-8 encoding should tell the system the character's width. It appears there is some automatic conversion from UTF-8 to UTF-32 (hence the wchar_t), but if this is the case, why is the conversion necessary?

Answer

ecatmur picture ecatmur · Mar 18, 2013

You use wchar_t because you're reading the file using wifstream; if you were reading using ifstream you'd use char, and similarly for char16_t and char32_t.

Assuming (as the example does) that wchar_t is 32-bit, and that the native character set that it represents is UTF-32 (UCS-4), then this is the simplest way to read a file as UTF-32; it is presented as such in the example for contrast to reading a file as UTF-16. A more portable method would be to use basic_ifstream<char32_t> and std::codecvt_utf8<char32_t> explicitly, as this is guaranteed to convert from a UTF-8 input stream to UTF-32 elements.