rounding zero with round-mode=figures

[quantenschaum]

The following code produces unexpected result

\sisetup{round-precision=3,round-mode=figures}
actual \num{0.000000} \num{1.000000} -- expected \num{0.00} \num{1.00}

renders as

There are too many trailing 0s when 0 is rounded. Using round-mode=places works as expected.

[josephwright]

I'm not sure what the 'expected' output is here: arguably it's undefined. I can for example see an argument for 0 (dropping all digits) as well as for the current behaviour. I'm not at all sure about the output you suggest: how do you decide it's 'right'?

[quantenschaum]

Well, I did not decide that it is right or wrong, I said it is "unexpected". Why? Simply because the 1 is rendered as 1.00 with 3 figures and I expected the 0 to be rendered with 3 figures (or maybe less, but not more than 3 as requested) as well. Trailing zeroes are removed from the 1 but not from the 0. Why?

And, when round-precision=2,round-mode=places is used, it behaves as expected. It renders 0.00 1.00.

I can imagine, that other users stumble upon this issue and the majority of them shares my expectation.

[josephwright]

What's important here is we are rounding to significant figures. So we get 1.00 because the 1 is a significant figure. If the input was 0.01000, we'd get 0.0100 as output. In the case 0.00000, there are no significant figures at all. So the output is really undefined: like I said, I can see an argument for 'no change' or `just zero'.

[josephwright]

Perhaps related: #206