printf: Default left-justify integer conversion to 1 width

[hbina]

When using left-justify with integer conversion (like printf '%-o'),
default the minimum width to 1.

Closes: #3050

Signed-off-by: Hanif Ariffin hanif.ariffin.4326@gmail.com

[tertsdiepraam]

The logic of the SubParser code is a bit hard to follow. Do you know if there is a possibility where min_width_tmp is not just empty but an invalid integer? Is that a case that we have to account for?

[sylvestre]

@hbina ping?

[hbina]

I don't fully understand the Output Conversion Syntax but I think what's happening is this:

The syntax for the conversion is something like this

% [ param-no $] flags width [ . precision ] type conversion

Given the input %-10o, the parser tokenize -10 as a token for the width.
It then converts the input to isize since it can be negative.
However, the previous author didn't consider the fact that %-o is also perfectly valid, and it just means that there is no minimum width required.
The default is 1 because AFAIK, for octals if there are no argument provided to it, it will default to 0.

I am not good enough at printf to try out different combination to break this so I might be totally wrong.

Do you know if there is a possibility where min_width_tmp is not just empty but an invalid integer? Is that a case that we have to account for?

GNU says that if the conversion syntax is ill-formed, the behavior is simply unpredictable so I don't think we have to care about invalid input.

If the syntax of a conversion specification is invalid, unpredictable things will happen, so don’t do this. If there aren’t enough function arguments provided to supply values for all the conversion specifications in the template string, or if the arguments are not of the correct types, the results are unpredictable. If you supply more arguments than conversion specifications, the extra argument values are simply ignored; this is sometimes useful.

Found at the bottom of the manual.