Ensuring integer are written as 32 bits

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AskBot

Question subject: Ensuring integer are written as 32 bits

Posted: Sat Nov 08, 2008 1:28 am

Joined: Sun Oct 19, 2008 3:47 pm
Posts: 281
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I have a program that writes a binary config file for another piece
of hardware. One of the requirements of the file is that integer
values are 32 bits. Currently the file is created as expected when
run on a 32 bit machine. However, I need to make it portable, so
that it could also be run on a machine that uses 64 bit ints, and
I'm not sure how to do it.

Currently, the code is something like this:

Code:

struct Data
{
int nInt1;
int nInt2;
char czString[32];
};

...

Code:

FILE* pFile = fopen("filename.bin", "wb");
Data data;
// do stuff

fwrite(&data, sizeof(data), 1, pFile);

...

Now, I know this is not portable because of int size (as well as
possible packing issues) so I thought about changing it to something
like this:

Code:

const int INT_SIZE = 4;
const int STRING_LENGTH = 32;
struct Data
{
int nInt1;
int nInt2;
char czString[STRING_LENGTH];
};

...

FILE* pFile = fopen("filename.bin", "wb");
Data data;
// do stuff

Code:

fwrite(&data.nInt1, INT_SIZE, 1, pFile);
fwrite(&data.nInt2, INT_SIZE, 1, pFile);
fwrite(&data.czString, 1, STRING_LENGTH, pFile);

...

Will this actually work? Will this write the lower 32 bits of an
int if it is 64 bits? If not, how can I guarantee it will write the
lower 32 bits?

AnswerBot

Question subject: Re: Ensuring integer are written as 32 bits

Posted: Sat Nov 08, 2008 1:35 am

Joined: Sun Oct 19, 2008 3:53 pm
Posts: 229
Has thanked: 0 time
Have thanks: 0 time

A 32-bit integer can be represented as an array of 4 chars,
assuming 8 bits per byte. Conversion between ordinary
integers and this array can be done by bitwise
mask-and-shift operations. The following should work for any
unsigned integer type >= 32 bits; I've used long long here:

Code:

#include <stdio.h>
#include <limits.h>

int main(void)
{
unsigned long long i = 1234567890;
unsigned char c[4];

printf("Your integers are %i bits\n", CHAR_BIT *
sizeof(i));

c[0] = i & 0xff; /* least significant byte */
c[1] = (i & 0xff00) >> 8;
c[2] = (i & 0xff0000) >> 16;
c[3] = (i & 0xff000000) >> 24;

printf("0x%08x == ", i);
printf("0x%02x%02x%02x%02x", c[3], c[2], c[1], c[0]);

return 0;
}

On my Win32 system this prints

Your integers are 64 bits
0x499602d2 == 0x499602d2

The first hex value is derived from i, the second from c[],
and they should be the same. I don't have a 64-bit machine
to try this on; maybe someone who has would oblige!

(To write binary data from c[] you'd need a different format
string of course.)

There are also considerations of endianess to be taken into
account.

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