Hi Wayne,
If i had
?- A :: 0.0 .. 0.5, [B, C] :: 0.0 .. 1.0, A $= eval(B + C).
A = A{0.0 .. 0.5}
B = B{0.0 .. 0.5}
C = C{0.0 .. 0.5}
There is 1 delayed goal.
Yes (0.00s cpu)
-- Professor Mark Wallace Room 6.43 Faculty of Information Technology Building H Monash University Caulfield Tel: 34276Received on Mon Jul 27 2009 - 11:24:47 CEST
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