On Tue, 24 Jul 2007, B. Bogart wrote:
Ah, I see now. I could not use that method, since 8/6 was an example, and I wanted to do $8 / $6, so treating "$8" as a symbol would not have helped.
Ah, caution with that: literals like 8 6 8.0 6.0 and pd's substitutors ($1 $2 $3) work in the same way in this case, but [expr]'s own substitutors ($f1 $f2 $f3) don't have that int-vs-float confusion problem because then [expr] does not attempt to read those floats as being text.
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