Re: A bug in parsing of conditional ops?

[arnold]

Denys Vlasenko <dvlasenk@redhat.com> wrote:

> On 7/9/24 16:04, arnold@skeeve.com wrote:
> >> since this unexpectedly works:
> >>
> >>    $ awk 'BEGIN { print 3==v=3, v}'
> >>    1 3
> >>
> >> According to the table, it should not, (3==v)=3 is not a valid assignment.
> > 
> > It's parsed (as you noted) as 3 == (v = 3)
>
> Well, it shouldn't. The == has higher precedence than =.

As I said, I don't really wish to dive into the grammar to try
to fix this.

However, if you wish to submit a patch that does, and doesn't bring
anything else, I will review it.

Thanks,

Arnold