从 char 向量转换为 short 向量和 float 向量

Question

I have written this program to convert from a char to a short:

typedef unsigned char u8;
typedef unsigned short u16;

u16 combine2bytes(u8 a, u8 b) {
    return b | (a << 8);
}


int main()
{
    std::vector<char> cVec{ 'a','r','t','u' };
    std::vector<short>sVec;
    for (auto i = 0; i < cVec.size(); i+=2)
    {
        sVec.push_back(combine2bytes(cVec.at(i),cVec.at(i+1)));
    }
    std::cout << sVec.size() << '\n';
    for (auto i = 0; i < sVec.size(); i++)
        std::cout << sVec.at(i) << '\n';
    return 0;
}

So if I understand, if we have a vector (or an array) of 1 byte data and I want to create a 2-byte vector from that 1-byte vector, I have to read 2 bytes from that 1-byte vector? Is this correct? same goes to float and int; we have to read 4 consecutive bytes?

but if I have to read from file, suppose this is the file:

F0 5A 6B A9 4C E3 etc..

suppose now that I want to cast it to short, so I have to take: F0 and 5A but which one is the LSB and MSB (NB: the structure is little Endian)?

And if I have to cast it to a float: how I can cast it so I can get floating point? because if I will cast it to a float, it's similar as I cast it to int? how to get the floating point?

Thanks in advance

Answer 1

The famous answer it depends fits here.

1. MSB/LSB : It depends on how you stored those bytes, for example if you store the MSB in vector case 0 then when reading you should consider this. You can not guess if you do not have this information.

2. data type : It depends on what the data you stored there. For example if it's a float then just do the bit shifing and then cast, you can try that by a concrete example. However, if it's another program who has written those bytes or they're not written on the same architecture, then you should have a protocol that defines which bytes are for the mantissa. again you can not guess if you do not have this information.

3. you code snippet: you should cast the data to short before bit-shifting and ORing it to ensure these operations are performed on 16-bits not on 8-bits data:

    u16 combine2bytes(u8 a, u8 b) { return ((u16)b) | (((u16)a) << 8); }