Running "crontab" with no options runs "crontab /dev/stdin"

[tometzky]

su username -c crontab -l
What this command does?

It does not list crontabs for username - for this it should be su username -c "crontab -l". It runs "crontab" with no options.

But what bare "crontab" command does? It's not documented - man page says at least one option or a single non-option argument is required. So I'd expect it to just fail with an error. But it runs an equivalent of "crontab /dev/stdin" instead, deleting the current crontab if stdin is empty. The "crontab -" also works like this (also undocumented in the man page).

The help showing when invalid option is given states "Default operation is replace, per 1003.2". I don't know the exact wording of the standard, but it does not seem to force this non-intuitive and dangerous behaviour. IMHO it just means that when the file argument is provided then the default operation is to replace the crontab with its contents.

[t8m]

This stupid behavior is specified by POSIX, but let's see if anyone complains if we drop such error prone behavior.

Fixed by 39a67fd

[mandree]

This "stupid behaviour" (i. e. require crontab - instead of crontab to replace the user's scheduled commands) that was removed makes cronie incompatible with other applications that expect POSIX behavior, and I fail to see why incompatibilities need to be introduced just because users can't be bothered to read and understand the manual.

A middle ground would be to use isatty() when crontab is called without arguments that prints something like crontab of user JOHN will be replaced, press ^C to abort and run crontab -l to print or something.

[t8m]

I've implemented a different but IMO still sufficient middle ground of requiring the - argument only if isatty(STDIN_FILENO).