Re: A bug in parsing of conditional ops?

[Denys Vlasenko]

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 =.


> >         $ awk 'BEGIN { print v=3==3, v}'
> >         1 1
>
> This is parsed as v = (3 == 3). C parses it the same way.

Yes. This one is correct, no argument here.
I am showing that gawk correctly handles precedence of =
and == in this case.


> > But there is more: more than one comparison op fails to parse:
> >
> >         $ awk 'BEGIN { print 3==3==3}'
> >         awk: cmd. line:1: BEGIN { print 3==3==3}
> >         awk: cmd. line:1:                   ^ syntax error
> >
> >         $ awk 'BEGIN { print 3==3!=3}'
> >         awk: cmd. line:1: BEGIN { print 3==3!=3}
> >         awk: cmd. line:1:                   ^ syntax error
>
> Brian Kernighan's awk treats it as a syntax error. Mawk
> actually prints 0 for the first one and 1 for the second.

Looks like mawk gives correct results:
"3==3" is 1, and "1==3" is 0
"3==3" is 1, and "1!=3" is 1