使用printf％q传递带空格的参数

Question

Suppose I have a script that prints out the number of arguments passed to it:

# file: num_args

echo "Number of arguments: $#"

Now my question pertains to the following invocations:

> ./num_args a b c
> Number of arguments: 3 # As I would expect.

> ./num_args "a b c"
> Number of arguments: 1 # As I would expect.

> ./num_args a\ b\ c
> Number of arguments: 1 # As I would expect.

> printf "%q\n" "a b c"
> a\ b\ c                # As I would expect.

> ./num_args $(printf "%q" "a b c")
> Number of arguments: 3 # NOT as I would expect.

Given that the printf man page states that

 %q     ARGUMENT is printed in a format that can be reused as shell input, escaping non-printable characters with the proposed POSIX $'' syntax.

I am not sure what happens in the last case above.

Answer 1

printf %q output is escaped correctly to be parsed as source code by a shell interpreter .

However, unquoted expansions aren't parsed as source code; they go only through string-splitting and glob-expansion phases. (This is a Good Thing: Otherwise handling untrusted filenames or other potentially-dangerous content in shell scripts would be nearly impossible).

You can substitute this output into a shell command, as in sh -c "...$foo..." or eval "...$foo..." or ssh somehost "...$foo..." ; it can't (shouldn't) just be used unquoted.

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Some important references:

• BashFAQ #50 explains why strings containing source code can't be used by expanding them unquoted.
• BashFAQ #48 explains the risks involved in using eval (one of the families of workarounds discussed above).