Declaring an array of bytes in c

[namratha]

We want to declare an array of 16 hexa decimal numbers of 8 bytes each.Like the following:

00000000_00000000
FFFFFFFF_FFFFFFFF
00000000_FFFFFFFF
0000FFFF_0000FFFF
00FF00FF_00FF00FF
0F0F0F0F_0F0F0F0F
33333333_33333333
55555555_55555555
AAAAAAAA_AAAAAAAA
CCCCCCCC_CCCCCCCC
F0F0F0F0_F0F0F0F0
FF00FF00_FF00FF00
FFFF0000_FFFF0000
FFFFFFFF_00000000
FFFFFFFF_FFFFFFFF
00000000_00000000

How can we achieve this in c?Any help is appreciated.

Thanks,
Namratha.

[weaknessforcats]

A hexadecimal digit requires 4 bits: F == 1111.

An unsigned integer is 32 bits, enough for 8 hex digits.
You number:

FFFFFFFF_FFFFFFFF


could be a struct variable:

struct Hex
{
   unsigned int left;            //big enough for FFFFFFFF
   unsigned int right;          //big enough for FFFFFFFF
};
 

I would supply the underscore as part of the displayu of these numbers.

Now you can write functions using variables of this Hex struct:

Hex Add(Hex left, Hex right); //returns left + right).

Or, you could use a string:

char anumber[18] = "FFFFFFFF_FFFFFFFF";
 

My preference would be to use the struct with two unsigned int members.

If you are using a 64 bit Unix system, a single unisgned int will do. However, if you ever need to port your code to 64-bit Windows, this approach won't work since 64-bit Windows has a 32-bit int. The 64-bit int is _int64. You might fudge this by using a macro: