Convert a 16-bit integer to an array of char? (C++)

Question

I need to write 16-bit integers to a file. fstream only writes characters. Thus I need to convert the integers to char - the actual integer, not the character representing the integer (i.e. 0 should be 0x00, not 0x30) I tried the following:

char * chararray = (char*)(&the_int);

However this creates a backwards array of two characters. The individual characters are not flipped, but the order of the characters is. Thus I created this function:

   char * inttochar(uint16_t input)
{
    int input_size = sizeof(input);
    char * chararray = (char*)(&input);
    char * output;
    output[0]='\0';
    for (int i=0; i<input_size; i++)
    {
        output[i]=chararray[input_size-(i+1)];
    }
    return output;
}

This seems slow. Surely there is a more efficient, less hacky way to convert it?

You can simply do f << i where f is your ofstream and i is your integer. If you want hex than you can do f << "0x" << std::hex << i. — Cheers and hth. - Alf
– Cheers and hth. - Alf, Commented Jan 23, 2016 at 15:41
I need to convert it to the the actual integer, not the character representing the integer. That method will turn an integer into the character representing the integer. — Nick
– Nick, Commented Jan 23, 2016 at 15:42
@Cheersandhth.-Alf The OP wants to write binary, big-endian (or to be precise, the endianness opposite of that native to their machine). Your proposal would produce text. — Igor Tandetnik
– Igor Tandetnik, Commented Jan 23, 2016 at 15:42
A bit more compact: char arr[2] = {the_int / 256, the_int % 256}; stream.write(arr, 2); — Igor Tandetnik
– Igor Tandetnik, Commented Jan 23, 2016 at 15:46

Ami Tavory · Accepted Answer · 2016-01-23 16:16:11Z

2

It's a bit hard to understand what you're asking here (perhaps it's just me, although I gather the commentators thought so too).

You write

fstream only writes characters

That's true, but doesn't necessarily mean you need to create a character array explicitly.

E.g., if you have an fstream object f (opened in binary mode), you can use the write method:

uint16_t s;
...
f.write(static_cast<const char *>(&s), sizeof(uint16_t));

As others have noted, when you serialize numbers, it often pays to use a commonly-accepted ordering. Hence, use htons (refer to the documentation for your OS's library):

uint16_t s;
...
const uint16_t ns = htons(s);
f.write(static_cast<const char *>(&ns), sizeof(uint16_t));

edited Jan 23, 2016 at 16:16

answered Jan 23, 2016 at 15:54

Ami Tavory

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1 Comment

Nick Over a year ago

Oh, by the way - in case anyone else has this problem - I don't think the static cast there works (it doesn't for me at least.) Maybe it should be changed to reinterpret_cast or just (char*)(&ns)?

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