bash：如何将十六进制整数打印为特定长度

Question

I am trying to dump the decimal integer values from one file in a hex format. I do have a file with integer values in decimal.

$ more test.dat_trim
2 9
0 -11
7 -17
14 -1

I am trying to print this integer in hex. I know also that the integer values are small enough to fit on 2 bytes. I want the output to be on 2 bytes. But then when i am trying:

declare -i i;for i in $(<test.dat_trim);do printf "%.2x\n" $i; done;
02
09
00
fffffffffffffff5
07
ffffffffffffffef
0e
ffffffffffffffff

Basically printf "%.2x\\n" it is only working for positive number. How can i make it work for negative also? Just to clarify what i am expecting: The result should be like this:

02
09
00
f5
07
ef
0e
ff

meaning that i want for the negative values to be sign extended only on 1 byte.

Answer 1

Printing signed hex values is uncommon, so there is no conversion specifier providing this.

You could use the following work around:

for i in $(<test.dat_trim); do
  if [ $i -ge 0 ]; then
    printf " 0x%02x\n" $i; 
  else
    printf "%c0x%02x\n" '-' $[$i * -1]; 
  fi
done;

---

Referrig the update to the question:

Just replace this line

printf "%c0x%02x\n" '-' $[$i * -1];

with this

printf " 0x%02x\n" $[256 + $i];

This however, only works for the numbers >= -256.

Answer 2

It can be done in awk, that handles negative numbers also:

awk '{printf "0x%x%s0x%x\n", $1, OFS, $2}' OFS='\t' file
0x2     0x9
0x0     0xfffffff5
0x7     0xffffffef
0xe     0xffffffff

Answer 3

金达傻了，但到底：

xargs -a test.dat_trim bash -c 'printf %.2s\\n $(printf %02x\\n $* | rev) | rev' _

Answer 4

您是否尝试过printf（“％04x \\ n”，i）？