Thanks for your comments and questions. 0. The C code presented is only for a primitive datatype (in C). The paper (script) says this. The basic hashing logic still works for other datatypes, but instead of using == and != (the C equals and not-equal) you'd need something more elaborate. Still simple conceptually because in your head you can subsume the details by thinking "match" or "not match".
1. The APL\360 defn assumes that the left argument is a vector. The "anomaly" you found only occurs if the left argument is not a vector. 2. *z++=*h<0?xn:*h; is equivalent to: blah= *h<0 ? xn : *h; // if *h is less than 0 then xn, otherwise *h *z=blah; // set the location pointed to by z, to blah z=z+1; // increment the pointer z by 1 This is standard C. In that context the statement is also equivalent to z[i]=*h<0?xn:*h; On Wed, Jul 30, 2014 at 8:53 AM, Brian Schott <[email protected]> wrote: > Roger, > > Because my C is so weak I could use some explanation in words of what the C > is doing, not in general, but to discuss the questions below. > > Would you say that the basic approach that is being taken in DO's is more > like the APL\360 definition or more like the Extended definition. I guess I > am a little surprised that a Loop is required through both the x and the y > values. In the Dictionary entry for Index of is the following example, > where the y component is not an atom, but a string. Is that a typical case > for which y must be looped through as well as x? I thought maybe the answer > was yes, and tried. > > A=: 'abcdefghijklmnopqrstuvwxyz' > m=: 5 4 $ 12{. A > m;(m i. 'efgh');(1{m);(4{m) > > I then found a slight "anomaly" though, trying the APL\360 definition with > that example. > > > +/*./\m~:/'efgh' > 1 5 5 5 > 5 1 5 5 > 5 5 1 5 > 5 5 5 1 > > However that "anomaly" was fixed by the Extended definition, so maybe > therein is an example of why the y value must be looped through, too. > > +/*./\m~:/&:<"(_1+#$m)'efgh' > 1 > > Can you comment, please? > > Also, in the second DO there is a phrase like this: *z++=*h<0?xn:*h; > > I am familiar with the B?T:F part of the phrase, but I have never seen a ++ > followed by an assignment. Does that just add one to the result from the > assigned amount? > > Oh, and thanks for posting your paper. > > > -- > (B=) > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
