The iterator doesn't change the algorithm (with assumption)

Post #128,844 by Simon_Jester 12/4/03 9:47:03 AM Reply	The iterator doesn't change the algorithm (with assumption) (Assumption: Smalltalk iterators work similar to Java && C++ iterators) Because you're walking through the array in both cases. 1st example (with iterator) collection collect collection ub collection lb Create 3 iterators, C = collect.iterator U = ub.iterator L = lb.iterator \nfor each result X {\n\n for (C = collect.begin(), U = ub.begin(), L = lb.begin();\n C != collect.end();\n C++, U++, L++) {\n if ((X < U.value()) && (X > L.value()) (C.value())++;\n }\n}\n Hint, it's even easier in method two, since you only need one iterator in this case. \nit = array.iterator();\nfor each result X {\n for (it = array.begin(); it != array.end(); it++) {\n if ((X < it.value().U) && (X > it.value().L)) (it.value().C)++;\n }\n}\n
Post #128,869 by Arkadiy 12/4/03 10:50:10 AM Reply	Smalltalk iterators are different You pass a block of code to a collection's method, and this block is executed for every array element: myArray do: [ :element\| element modifyAsNeeded] A variation I used in my cheating implementation was: randomNumbers groupedBy: [:number \| (number / increment) truncated] This produces a dictionary of sets that has integers as keys, and set of all numbers that produced that integer in the block as vaules. -- The rich, as usual, are employing the elected. -- [link\|http://unfit2print.blogspot.com/\|http://unfit2print.blogspot.com/]
Post #128,884 by Simon_Jester 12/4/03 11:44:55 AM Reply	Okay...so your 'cheat' is the method Rather than walk through our collection, we apply to each element of the collection at one time. I still would use my 2nd way, making the collection a set of objects, each object knowing whether or not to increment it's counter when presented with a random number X. (Each object would have an upper/lower bound and test for them). Add a method to pull out the counter out of each object afterwards and you're done.
Post #128,891 by Arkadiy 12/4/03 11:59:42 AM Reply	see my answer to Ben below... -- The rich, as usual, are employing the elected. -- [link\|http://unfit2print.blogspot.com/\|http://unfit2print.blogspot.com/]

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