print a file in hex using awk

Question

I have a file name x

$ cat x
1A34532112345

I wants to print in hex using awk I written a script (with help from web)

fold -1 /home/cscape/Desktop/x | gawk '{ printf("%s , %X\n",$0, int($0) )}' 

But output for each char is 0

$ fold -1 /home/cscape/Desktop/x | gawk '{ printf("%s , %X\n",$0, int($0) )}'
1 , 1
A , 0
3 , 3
4 , 4
5 , 5
3 , 3
2 , 2
1 , 1
1 , 1
2 , 2
3 , 3
4 , 4
5 , 5

Why 0 for A. even with %d. I wants to print ASCII value of A as HEX.

Answer 1

awk will silently convert strings to numbers. Which number? The number you get by taking the initial decimal digits of the string (this may be just "", if the string is empty or doesn't begin with any decimal digits, as is the case with "A") and converting them to a number. "" gets converted to 0. So all of these would be converted to 0:

• "0"
• "0text"
• "text"
• ""

awk only converts strings to numbers in this way when a number is needed. In "0text" + 1, a number is needed, so the result will be 1. In just plain "0text", a number isn't needed, so no conversion takes place. In printf("%d", "0text"), a number again is needed, so the string will be converted to a number.

What you are looking for is an ord function, which isn't a native function in awk. The gawk documentation describes how to write such a function in awk.

Answer 2

If you don't have to use awk, you might look at od ("octal dump"):

$ echo 1A34532112345 | od -t x1
0000000 31 41 33 34 35 33 32 31 31 32 33 34 35 0a
0000016