eqv?, to be exact. Your case statement works fine for me, with for example (comp->text '=). (eq? comp =) compares against the value of the procedure =, whereas the case compares against the symbol =. So you are doing two different comparisons.
You can use ,x (case ...) at the REPL to see what the case expands to. Fake edit: Kon said this more succinctly than I did. Jim On Mar 2, 2012, at 5:52 PM, Matt Welland wrote: > I expected this to work: > > (define (comp->text comp) > (case comp > ((=) "=") > ((>) ">") > ((<) "<") > ((>=) ">=") > ((<=) "<=") > (else "unk"))) > > But had to convert to a cond with (eq? comp =) etc... > > I thought case was supposed to use eq? to do the compare?
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