Re: How to safely print a variable with secret value to stdout?

[Greg Wooledge]

On Thu, May 12, 2022 at 02:36:20AM +0200, Philippe Cerfon wrote:
> printf '%s' "$secret" | foo --password-from-stdin
> 
> should be safe and not show up in the process list or so.
> 
> The problem is that the code might be run on shells where this isn't
> guaranteed

You control the shell with the shebang.  If you use #!/bin/bash or
whatever shebang is appropriate to invoke bash on the target system,
then you know you're using a builtin printf.

That said,

https://mywiki.wooledge.org/BashFAQ/069