Copy a std::u8string into a c-style string of utf8 characters

Question

Copying a string with no encodage into a c-string is quite easy:

auto to_c_str(std::string const& str) -> char* {
    auto dest = new char[str.size() + 1];
    return strcpy(dest, str.c_str());
}

But how can I do that with a std::u8string? Is there a STL algorithm that can help with that?

I tried this:

auto to_c_str(std::u8string const& str) -> char8_t* {
    auto dest = new char8_t[str.size() + 1];
    return std::strcpy(dest, str.c_str());
}

But of course, std::strcpy is not overloaded for utf8 strings.

The way UTF-8 is defined, I don't see any problem with what you have. Might as well just use `memcpy` for performance. Or am I missing the point? — DeiDei, Jul 02 '19 at 20:06

Ted Lyngmo · Answer 1 · 2019-07-02T21:20:53.937

10

strcpy isn't needed since you already know the length of what you'd like to copy, so use memcpy:

char8_t* to_c_str(std::u8string const& str) {
    auto dest = new char8_t[str.size() + 1];
    return static_cast<char8_t*>(std::memcpy(dest, str.data(), str.size()+1));
}

or std::copy:

char8_t* to_c_str(std::u8string const& str) {
    auto dest = new char8_t[str.size() + 1];
    std::copy(str.data(), str.data() + str.size() + 1, dest);
    return dest;
}

Since the u8string's own copy() method can't be used to include the null-terminator directly, I'd not use it when copying to a raw char8_t*.

edited Jul 02 '19 at 21:20

answered Jul 02 '19 at 20:07

Ted Lyngmo

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I never had to deal with a lot of c-style strings. This clears things up, thanks. – Guillaume Racicot Jul 02 '19 at 20:12

R Sahu · Accepted Answer · 2019-07-02T20:30:33.687

3

In addtion to using std::memcpy, you may use std::u8string::copy and std::copy.

auto to_c_str(std::u8string const& str) -> char8_t* {
    auto dest = new char8_t[str.size() + 1];
    str.copy(dest, str.size(), 0);
    dest[str.size()] = u8'\0';
    return dest;
}

auto to_c_str(std::u8string const& str) -> char8_t* {
    auto dest = new char8_t[str.size() + 1];
    std::copy(str.begin(), str.end(), dest);
    dest[str.size()] = u8'\0';
    return dest;
}

edited Jul 02 '19 at 20:30

answered Jul 02 '19 at 20:15

R Sahu

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Or `std::char_traits::copy` just to annoy ;) – DeiDei Jul 02 '19 at 20:17
@DeiDei I presume `std::u8string::copy` will use what char traits implements? – Guillaume Racicot Jul 02 '19 at 20:20
@DeiDei, who kew there were so many ways to copy a sequence of characters! – R Sahu Jul 02 '19 at 20:21
@GuillaumeRacicot Most probably, and in the end you'll end up with a `memcpy` somewhere. – DeiDei Jul 02 '19 at 20:21
1

Would the null terminator be `u8'\0'`? – Guillaume Racicot Jul 02 '19 at 20:23
@GuillaumeRacicot, I like that suggestion. Thanks. – R Sahu Jul 02 '19 at 20:30

score 1 · Answer 3 · answered Jul 02 '19 at 20:27

It seems to me like it would be easier to simply leverage the built-in copying and provide .data() to the C code:

std::u8string orig = u8"abc";
auto copy = orig;
c_api(copy.data(), copy.size());

By doing this, you let the copied string manage its own lifetime and have the size on equal footing with the data. This works uniformly for any char type of std::basic_string. As an added bonus, it also works for std::vector.

Copy a std::u8string into a c-style string of utf8 characters

3 Answers3