Function to convert an int to hex

[o.fithcheallaig]

Hello.

What are you looking to do when writing a function that will convert an into to it's hex version?

I think I am right in saying I need to keep dividing the number by 16, until I get a remained of zero, yea? And the results will be the answer. Of course, I need to take into account the A-F. But Would this be done in a for loop?

Seán

[MK27]

I think you mean, convert a decimal number to a hexadecimal number. Ints are neither; they are stored as binary. In source code, by default numbers are presumed to be decimal, but you can also use hex via 0x.

If you want a string representation of an int in hexadecimal:

	unsigned int n = 0xdeadbeef;
	char hex[sizeof(int)*2 + 1];
	sprintf(hex, "%x", n);
	puts(hex);
	printf("%u\n", n);

You don't have to use an unsigned type, I just like to say "deadbeef" and it won't fit otherwise.

[nonoob]

Yes, o.fithcheallaig. Divide by 16, and each time you get a remainder which strips off the last hex digit. That way you are constructing the result from least-significant-digit to most-significant-digit. Indicates you save the digits in an array and display it out backwards. You can use a string such as "0123456789ABCDEF" to index into and display digits.