I am trying to printf some unicode codes that I pipe in like this

echo 0024 0025 | xargs -n1 echo # one code per line
  | xargs printf '\u%s\n'

hoping to get this


but this is what I get

printf: missing hexadecimal number in escape

After some trial and error, I actually have two smaller problems, and one kind-of makes sense and the other seems like a complete mystery.

Problem 1:

printf '\u%s\n' 0024 0025

gives me this

-bash: printf: missing unicode digit for \u
-bash: printf: missing unicode digit for \u

Problem 2:

> # use built-in for $
> printf '\u0024\n'
> # use exe for $
> which printf
> /usr/bin/printf '\u0024\n'
> # now use built-in for %
> printf '\u0025\n'
> # but look what happens when we use exe for % !!!!
> /usr/bin/printf '\u0025\n'
/usr/bin/printf: invalid universal character name \u0025

(using > for $ so you can see the $ in the output)

For some reason some characters work with exe version but some don't even though all work with built-in printf.

so here is a work-around that would work if it weren't for problem #2 (but might be quite a bit slower than my original idea)

echo 0024 0025 | xargs -n1 echo # one item per line
  | xargs -I {} printf '\u{}\n'

but due to problem #2, it kind of half works:

$ echo 0024 0025 | xargs -n1 echo | xargs -I {} printf '\u{}\n'
printf: invalid universal character name \u0025

($ comes out but % gets error)

So I guess my questions are:

-Is there any way of making printf work with the number code so that I can run printf once instead of once per argument with -I?

-What am I doing wrong that printf built-in doesn't mind, but printf exe doesn't like, but only for % and not for $?

  • 1
    Add an additional \` e.g. printf '\\u%s\n' 0024 0025`
    – Jetchisel
    Apr 14, 2020 at 11:51
  • 1
    I just get something like \u0024
    – Alex028502
    Apr 14, 2020 at 13:00
  • Instead of struggling with chains of commands using unportable GNU extensions, just use perl perl -CO -nle 'print chr hex for /[0-9a-fA-F]+/g' and be done with it.
    – user313992
    Apr 15, 2020 at 3:45

2 Answers 2


To avoid the double-expansion problem (\u is processed before %s), you can use %b, at least in Bash printf:

printf '%b\n' \\u0024 \\u0025

You can pre-process your input in various ways:

set 0024 0025
printf '%b\n' "${@/#/\\u}"

The standalone printf, as implemented in GNU coreutils, has the following restrictions on Unicode character specifications:

printf interprets two character syntaxes introduced in ISO C 99: ‘\u’ for 16-bit Unicode (ISO/IEC 10646) characters, specified as four hexadecimal digits hhhh, and ‘\U’ for 32-bit Unicode characters, specified as eight hexadecimal digits hhhhhhhh. printf outputs the Unicode characters according to the LC_CTYPE locale. Unicode characters in the ranges U+0000…U+009F, U+D800…U+DFFF cannot be specified by this syntax, except for U+0024 ($), U+0040 (@), and U+0060 (`).

This explains why you can’t produce % in this manner.


The standard printf utility does not support \uxxxx escape sequences, see: https://pubs.opengroup.org/onlinepubs/9699919799/utilities/printf.html

Assuming this could work depends on extensions that might be present in a few implementations (e.g. in a ksh builtin), but cannot be expected to be supported in general. See the printf standard document.

Another problem seems to be your assumption that calling

printf '\u%s\n' 123

would result in the same as calling:

printf '\u123\n'

This does not work, as printf parses the format string element by element and does not see the expected format string.

So even if you are using bash to execute the script, you may just expect the \uxx backslash escape to be expanded if two hex digits follow and the escape sequence appears literally in the format string. If you like to have 4 hex digits expanded, you need to have \Uxxxx literally in the format string.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .