Question

I know that to get a unicode character in C++ I can do:

std::wstring str = L"\u4FF0";

However, what if I want to get all the characters in the range 4FF0 to 5FF0? Is it possible to dynamically build a unicode character? What I have in mind is something like this pseudo-code:

for (int i = 20464; i < 24560; i++ { // From 4FF0 to 5FF0
    std::wstring str = L"\u" + hexa(i); // build the unicode character
    // do something with str
}

How would I do that in C++?

Answer 1

The wchar_t type held within a wstring is an integer type, so you can use it directly:

for (wchar_t c = 0x4ff0;  c <= 0x5ff0;  ++c) {
    std::wstring str(1, c);
    // do something with str
}

Be careful trying to do this with characters above 0xffff, since depending on the platform (e.g. Windows) they will not fit into a wchar_t.

If for example you wanted to see the Emoticon block in a string, you can create surrogate pairs:

std::wstring str;
for (int c = 0x1f600; c <= 0x1f64f; ++c) {
    if (c <= 0xffff || sizeof(wchar_t) > 2)
        str.append(1, (wchar_t)c);
    else {
        str.append(1, (wchar_t)(0xd800 | ((c - 0x10000) >> 10)));
        str.append(1, (wchar_t)(0xdc00 | ((c - 0x10000) & 0x3ff)));
    }
}