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argument of with_key doesn't parse correctly #53
Description
Syntactic bug in implementation of #29.
This works:
pathto(start) min= 0 with_key [start].
pathto(V) min= pathto(U) + edge(U,V) with_key [V | bestpath(U)].
bestpath(U) = $key(pathto(U)).
But this doesn't:
pathto(start) min= 0 with_key [start].
pathto(V) min= pathto(U) + edge(U,V) with_key [V | $key(pathto(U))].
In the second case, the second rule never seems to fire. It may be misparsing the | as "or" rather than list syntax. (To understand the message below, note that the vertices in this graph are 5-letter words such as "black" and "blank".)
Error(s) in rule: <repl>
pathto(V) min= pathto(U) + edge(U,V) with_key [V | $key(pathto(U))].
TypeError:
when `$key(pathto("black"))` = ["black"]
pathto(V="blank") min= &with_key((pathto(U="black")=? + edge(U="black", V="blank")=10)=?, [(V="blank" | $key(&pathto(U="black"))=["black"])=?]).
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timvieira commentedon Jul 17, 2013
See also discussion on #15
timvieira commentedon Jul 17, 2013
Btw, In the mean time you use single quotes '$key' to side-step the bug.
jeisner commentedon Jul 18, 2013
Ah yes, problem already reported and diagnosed there. But you closed that ticket so I guess the issue will live on here.
nwf commentedon Jul 18, 2013
Er, wait, wasn't this fixed by ddb1042? Jason, can you tell me which version of dyna you're using? "(cd
which dyna; git describe --always)" is probably the best way to do that right now.jeisner commentedon Jul 18, 2013
f4d6dd0