thanks for your information 2010/3/22 Kan-Ru Chen <[email protected]>: > FYI, there is a tool called makedict can be used to create the binary format. > > On Mon, Mar 22, 2010 at 2:00 PM, a a <[email protected]> wrote: >> i'm sorry, there is get child count after get the child address. >> so it seam good. >> >> 2010/3/22 a a <[email protected]>: >>> Dear All, >>> I met a big difficiult problem, i have researched more then 5 >>> day's. Of course i read the LatinIME's algorithm for the suggestion. >>> It following like that >>> >>> getSuggestions-->getWordsRec, and getWordsRec invoke itself when it >>> have a child. So the key algorithm is where is it's child.And it is: >>> intDictionary::getAddress(int *pos) // the pos is a father node >>> { >>> int address = 0; >>> if ((mDict[*pos] & FLAG_ADDRESS_MASK) == 0) { //// The bit that >>> decides if an address follows in the next 22 bits >>> >>> // #define FLAG_ADDRESS_MASK 0x40 >>> >>> // 22-bit address = ~4MB dictionary size >>> limit, which on average would be about 200k-300k words >>> >>> *pos += 1; // so it looks never go into this branch when the >>> dictionay is less then 4MB, and my dictionary is only 500KB >>> } else { >>> address += (mDict[*pos] & (ADDRESS_MASK >> 16)) << 16; // >>> address += (mDict[*pos + 1] & 0xFF) << 8; >>> // >>> address += (mDict[*pos + 2] & 0xFF); >>> // >>> *pos += 3; >>> } >>> return address; //the address is child node's >>> } >>> For example, we typed "as", the first node is find the "a" at the root >>> layler, and then find "s" on the "a" child. >>> Anyone can tell me what's the getAddress() meanings? i really can't >>> understand it well. My question is why >>> the child address is always : >>> >>> address += (mDict[*pos] & (ADDRESS_MASK >> 16)) << 16; // >>> address += (mDict[*pos + 1] & 0xFF) << 8; >>> // >>> address += (mDict[*pos + 2] & 0xFF); >>> // >>> >>> If the "a" have more then two child, it will only get one child, >>> because it always get it's child address like above. >>> >> >> -- >> You received this message because you are subscribed to the Google >> Groups "Android Developers" group. >> To post to this group, send email to [email protected] >> To unsubscribe from this group, send email to >> [email protected] >> For more options, visit this group at >> http://groups.google.com/group/android-developers?hl=en >> >> To unsubscribe from this group, send email to >> android-developers+unsubscribegooglegroups.com or reply to this email with >> the words "REMOVE ME" as the subject. >> > > -- > You received this message because you are subscribed to the Google > Groups "Android Developers" group. > To post to this group, send email to [email protected] > To unsubscribe from this group, send email to > [email protected] > For more options, visit this group at > http://groups.google.com/group/android-developers?hl=en > > To unsubscribe from this group, send email to > android-developers+unsubscribegooglegroups.com or reply to this email with > the words "REMOVE ME" as the subject. >
-- You received this message because you are subscribed to the Google Groups "Android Developers" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/android-developers?hl=en To unsubscribe from this group, send email to android-developers+unsubscribegooglegroups.com or reply to this email with the words "REMOVE ME" as the subject.
