Q: How can I convert integers to binary or hexadecimal?
A: Make sure you really know what you're asking. Integers are stored internally in binary, although for most purposes it is not incorrect to think of them as being in octal, decimal, or hexadecimal, whichever is convenient. The base in which a number is expressed matters only when that number is read in from or written out to the outside world, either in the form of a source code constant or in the form of I/O performed by a program.
In source code, a non-decimal base is indicated by a leading 0 or 0x (for octal or hexadecimal, respectively). During I/O, the base of a formatted number is controlled in the printf and scanf family of functions by the choice of format specifier (%d, %o, %x, etc.) and in the strtol and strtoul functions by the third argument. During binary I/O, however, the base again becomes immaterial: if numbers are being read or written as individual bytes (typically with getc or putc), or as multi-byte words (typically with fread or fwrite), it is meaningless to ask what ``base'' they are in.
If what you need is formatted binary conversion, it's easy enough to do. Here is a little function for formatting a number in a requested base:
char *
baseconv(unsigned int num, int base)
{
static char retbuf[33];
char *p;
if(base < 2 || base > 16)
return NULL;
p = &retbuf[sizeof(retbuf)-1];
*p = '\0';
do {
*--p = "0123456789abcdef"[num % base];
num /= base;
} while(num != 0);
return p;
}
(Note that this function, as written,
returns a pointer to static data,
such that only one of its return values can be used at a time;
see question 7.5a.
A better size for the retbuf array
would be sizeof(int)*CHAR_BIT+1;
see question 12.21.)
For more information about ``binary'' I/O, see questions 2.11, 12.37, and 12.42. See also questions 8.6 and 13.1.
Additional links: A long reply I sent to someone who was asking how to write a ``binary to decimal'' conversion function
References:
ISO Secs. 7.10.1.5,7.10.1.6
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