Why use hex?

Question

Hey! I was looking at this code at http://www.gnu.org/software/m68hc11/examples/primes_8c-source.html

I noticed that in some situations they used hex numbers, like in line 134:

for (j = 1; val && j <= 0x80; j <<= 1, q++) 

Now why would they use the 0x80? I am not that good with hex but I found an online hex to decimal and it gave me 128 for 0x80.

Also before line 134, on line 114 they have this:

small_n = (n & 0xffff0000) == 0; 

The hex to decimal gave me 4294901760 for that hex number. So here in this line they are making a bit AND and comparing the result to 0??

Why not just use the number? Can anyone please explain and please do give examples of other situations.

Also I have seen large lines of code where it's just hex numbers and never really understood why :(

Answer 1

In both cases you cite, the bit pattern of the number is important, not the actual number.

For example, In the first case, j is going to be 1, then 2, 4, 8, 16, 32, 64 and finally 128 as the loop progresses.

In binary, that is,

0000:0001, 0000:0010, 0000:0100, 0000:1000, 0001:0000, 0010:0000, 0100:0000 and 1000:0000.

There's no option for binary constants in C (until C23) or C++ (until C++14), but it's a bit clearer in Hex: 0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, and 0x80.

In the second example, the goal was to remove the lower two bytes of the value. So given a value of 1,234,567,890 we want to end up with 1,234,567,168.
In hex, it's clearer: start with 0x4996:02d2, end with 0x4996:0000.

Answer 2

its a bit mask. Hex values make it easy to see the underlying binary representation. n & 0xffff0000 returns the top 16 bits of n. 0xffff0000 means "16 1s and 16 0s in binary"

0x80 means "1000000", so you start with "00000001" and continue shifting that bit over to the left "0000010", "0000100", etc until "1000000"