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I currently have this (pseudo) code

private:
    uint8_t* m_aBuf;

public:
    template <>
    auto read<std::string>() -> std::string
    {
        auto usLength = this->decode<uint16_t>();
        auto pStr = std::make_unique<char[]>(usLength);
        std::copy(this->m_aBuf, this->m_aBuf + usLength, &pStr.get()[0]);

        return std::string(pStr);
    }

Assuming the buffer looks like this:

05 00 68 65 6c 6c 6f

My expectation is for the specialized read<std::string>() to return a C++ string containing "hello". The size of the string is known during run time of the function (it's usLength).

(Yes I'm aware this will only work on little-endian systems and that's fine)

My intention is to avoid having the overhead of allocating the char[] array.

Can I do that? And how? Or is there a better solution for what I'm trying to achieve?

Improve this question
4
  • 1
    You could resize the std::string to the required length (std::string::resize()). Afterwards, you can copy the data into the std::string::data(). AFAIK, it is safe as long as you copy the appropriate size of chars. Commented Mar 23, 2020 at 14:29
  • 2
    FYI: SO: Is it permitted to modify the internal std::string buffer returned by operator[] in C++11 Commented Mar 23, 2020 at 14:32
  • 1
    @Scheff that worked by copying to &str.data()[0]. I'm surprised I couldn't use str.reserve() and i had to use str.resize() though but that's answered in stackoverflow.com/questions/9521629/… Commented Mar 23, 2020 at 14:33
  • 2
    std::string::reserve() is not the right tool for this. std::string::resize() grants that the internal buffer provides at least the required size (if it doesn't fail). So, the internal buffer of size must be there afterwards, which makes the little raw-overwrite-trick safe. (Even if reserve() would provide the memory for granted - it's still not claimed to be in use.) Commented Mar 23, 2020 at 14:36
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