Ralf Hemmecke wrote:
>
> Let x be of type SExpression. I would like to test whether x is the Lisp
> value T. I can do this via
>
> foo(x: SExpression): Boolean ==
> TRUE: SExpression := EQ(x,x)$Lisp
> x = TRUE
I am affraid this is one of simplest ways.
> But is there a simpler looking way of doing it. I had problems with
>
> bar(x: SExpression): Boolean ==
> TRUE: SExpression := T$Lisp
> x = TRUE
>
> because T$Lisp would denote T$ and not T.
We rename T to avoid accidental capture. Of course, this
gets into your way when capture is intentional...
--
Waldek Hebisch
[email protected]
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