Memcpy uint32_t into char*

Question

I testing a bit with different formats and stuff like that. And we got a task where we have to put uint32_t into char*. This is the code i use:

void appendString(string *s, uint32_t append){
    char data[4];
    memcpy(data, &append, sizeof(append));
    s->append(data);
}

void appendString(string *s, short append){
    char data[2];
    memcpy(data, &append, sizeof(append));
    s->append(data);
}

From string to char is simple and we have to add multiple uints into the char*. So now i'm just calling it like:

string s;
appendString(&s, (uint32_t)1152); //this works
appendString(&s, (uint32_t)640); //this also works
appendString(&s, (uint32_t)512); //this doesn't work

I absolutely don't understand why the last one isn't working properly. I've tested multiple variations of transform this. One way always gave me output like (in bits): 00110100 | 00110101 ... so the first 2 bits are always zero, followed by 11 and then for me some random numbers.. What am i doing wrong?

Answer 1

Assuming that string is std::string, then the single-argument version of std::string::append is being used, which assumes the input data is NUL-terminated. Yours is not, but append will go looking for the first NUL byte anyway.

512 is 0x00000100, which on a little endian machine is 0x00 0x01 0x00 0x00. Since the first byte is NUL, std::string::append() stops there.

Use the version of std::string::append() where you pass in the length.