On Sun, September 29, 2013 16:01, ZyX wrote:
> Try the following code:
>
> let d={'0d': 1}
> if 0 && (d.0d)
> endif
>
> . This code is completely correct if you replace `if 0` with `if 1`, but
> with `if 0` it throws
>
>
> Error detected while processing /home/zyx/tmp/vim/subscript-bug.vim:
> line 2:
> E110: Missing ')'
> E15: Invalid expression: 0 && (d.0d)
>
> . Reason is the following: when `evaluate` argument is set to FALSE in
> eval7() handle_subscript does not take .0d as dictionary key (because it
> checks for variable type which is VAR_UNKNOWN if evaluate is FALSE). Thus
> `.0d` is handled in eval5() like string concatenation and when it tries to
> evaluate second argument “0d” in eval7() it is handled like number in that
> big switch(**arg) statement. As a number it can consume only leading zero,
> which means when eval7() *that is handling top “(d.0d)” expression* sees
> that in place of getting **arg equal to ')' it gets **arg equal to 'd' and
> errors out.
This looks a little bit like the bug that has been fixed with 7.3.841.
Nevertheless the fix seems more complicated this time, as I don't know
how to only consume the argument for dictionaries while leaving string
concatenation alone.
regards,
Christian
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