R help - Sep 2003 - Another R syntax question

The file src/library/base/R/print.R
contains this line:

    x0 <- xm[okP]==0 != (as.numeric(Cf[okP])==0)

I didn't know R allowed that, and I wonder if it is deliberate?
                                        
In mathematics, you would expect x = y not= z to mean
(x = y) and (y not= z).;
In R, it is parsed as (x == y) != z.

While I have to admit that the spacing around == and != in print.R's
example does actually convey the actual association, in a vaguely
Principia Mathematica-ish way, I have to wonder:

Is R really seriously intended to swallow x == y != z without complaint?

If the line

%left       GT GE LT LE EQ NE

in src/main/gram.y were changed to

%nonassoc   GT GE LT LE EQ NE

then the line

    x0 <- xm[okP]==0 != (as.numeric(Cf[okP])==0)

in src/library/base/R/print.R would have to be changed to

    x0 <- (xm[okP]==0) != (as.numeric(Cf[okP])==0)

and all things considered, surely that would be an improvement?

On Wed, 3 Sep 2003, Richard A. O'Keefe wrote:
> The file src/library/base/R/print.R
> contains this line:
>
>     x0 <- xm[okP]==0 != (as.numeric(Cf[okP])==0)
>
> I didn't know R allowed that, and I wonder if it is deliberate?
Well, I'm not surprised that it's syntactically valid, but I
wouldn't know
off-hand which way it evaluated.
> In mathematics, you would expect x = y not= z to mean
> (x = y) and (y not= z).;
> In R, it is parsed as (x == y) != z.
I think it's clear that it will parse as either
(x==y) != z
or
x == (y!=z)
but not which.
> While I have to admit that the spacing around == and != in print.R's
> example does actually convey the actual association, in a vaguely
> Principia Mathematica-ish way, I have to wonder:
>
> Is R really seriously intended to swallow x == y != z without complaint?
>
> If the line
>
> %left       GT GE LT LE EQ NE
>
> in src/main/gram.y were changed to
>
> %nonassoc   GT GE LT LE EQ NE
>
> then the line
>
>     x0 <- xm[okP]==0 != (as.numeric(Cf[okP])==0)
>
> in src/library/base/R/print.R would have to be changed to
>
>     x0 <- (xm[okP]==0) != (as.numeric(Cf[okP])==0)
>
> and all things considered, surely that would be an improvement?

This is presumably S compatible (<checks: yes it is>) and so while
changing print.R would be an improvement, changing the grammar probably
wouldn't.

	-thomas