Look closer.

Code:

#include <stdio.h>
int main(void)
{
        int n=32;
        fwrite(&n, 1, sizeof(n), stdout);
}

Code:

$ ./a.out > bin.bin
$ hexdump -C bin.bin
00000000  20 00 00 00                                       | ...|
00000004

Imagine that; you get four binary bytes representing the integer 32. Because this is a little-endian system, they end up all backwards. Try echo $((0x00000020)) in your shell. It is. They are numbers. Binary numbers.

---------- Post updated at 09:14 AM ---------- Previous update was at 09:08 AM ----------

'int' types are 32-bit even on most 64-bit systems. long integers, though, are generally 64-bit on 64-bit systems (and 32-bit elsewhere). And if your system is neither 32 nor 64 bits, all bets are off.

If you're concerned about your integers changing size when your code gets moved, you can #include <stdint.h> and use int32_t to get a 32-bit integer that'll always be a 32-bit integer.