How to make calculations on hexadecimal numbers with awk?

Question

I have a file containing a list of hexadecimal numbers, as 0x12345678 one per line.

I want to make a calculation on them. For this, I thought of using awk. But if printing an hexadecimal number with awk is easy with the printf function, I haven't find a way to interpret the hexadecimal input other than as text (or 0, conversion to integer stops on the x).

awk '{ print $1; }'                     // 0x12345678
awk '{ printf("%x\n", $1)}'             // 0
awk '{ printf("%x\n", $1+1)}'           // 1                 // DarkDust answer
awk '{ printf("%s: %x\n", $1, $1)}'     // 0x12345678: 0

Is it possible to print, e.g. the value +1?

awk '{ printf(%x\n", ??????)}'          // 0x12345679

Edit: One liners on other languages welcomed! (if reasonable length ;-) )

Answer 1

gawk has the strtonum function:

% echo 0x12345678 | gawk '{ printf "%s: %x - %x\n", $1, $1, strtonum($1) }'
0x12345678: 0 - 12345678

Answer 2

In the original nawk and mawk implementations the hexadecimal (and octal) numbers are recognised. gawk (which I guess you are using) has the feature/bug of not doing this. It has a command line switch to get the behaviour you want: --non-decimal-data.

echo 0x12345678 | mawk '{ printf "%s: %x\n", $1, $1 }'
0x12345678: 12345678

echo 0x12345678 | gawk '{ printf "%s: %x\n", $1, $1 }'
0x12345678: 0

echo 0x12345678 | gawk --non-decimal-data '{ printf "%s: %x\n", $1, $1 }'
0x12345678: 12345678