bash printf + echo string

Question

So I'm trying to print a string as a argument to a program in the terminal. However, my command doesn't seem to match the string, which I need. The string I need to print is:

aaaaa$'\x14\x84\x04\x08'

My command:

`printf 'a%.0s' {1..76}; echo "$'\x14\x84\x04\x08'"`

Am I just making a stupid mistake?

When you used in the echo statement double-quotes surrounding the $'...' inhibited the expansion. — 0x2b3bfa0
– 0x2b3bfa0, Commented May 13, 2014 at 19:47

Adem Öztaş · Accepted Answer · 2014-05-13 14:29:41Z

3

Try like this,

~$ printf %76s |tr " " "a"; echo "$'\x14\x84\x04\x08'"
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa$'\x14\x84\x04\x08'

answered May 13, 2014 at 14:29

Adem Öztaş

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1 Comment

chepner Over a year ago

You can also dispense with the echo: printf '%76s\x14\x84\x04\x08' | tr ....

0x2b3bfa0 · Accepted Answer · 2014-05-13 16:00:57Z

0

¿Something like this?

 `printf "%76s"; echo "$'\x14\x84\x04\x08'"`

If do you want to print the hex characters:

 `printf "%76s"; echo $'\x14\x84\x04\x08'`

Is recommended to use the dollar sign and parentheses instead of backticks:

 $(printf "%76s"; echo "$'\x14\x84\x04\x08'")

Or

 $(printf "%76s"; echo $'\x14\x84\x04\x08')

edited May 13, 2014 at 16:00

answered May 13, 2014 at 15:53

0x2b3bfa0

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