@ritesh
umm, well here's a simple testcase to show the problem in the code......
isMatch("aa", "a*")On Thu, Dec 27, 2012 at 7:17 PM, Ritesh Mishra <rforr...@gmail.com> wrote: > @shady : either the string will be stored in heap or stack. thus accessing > address in heap or stack is not going to give u seg fault . and rest things > are very well handled in the code :) > As saurabh sir has explained in thread > https://mail.google.com/mail/u/1/#inbox/13ba918bdb9aac9e > when seg fault occurs . > Regards, > > Ritesh Kumar Mishra > Information Technology > Third Year Undergraduate > MNNIT Allahabad > > > On Thu, Dec 27, 2012 at 6:43 PM, ~*~VICKY~*~ <venkat.jun...@gmail.com>wrote: > >> I'm giving you a simple recursive code which i wrote long back. Please >> let me know if it fails for any cases. Ignore the funny cout's It used to >> help me debug and i'm lazy to remove it. :P :) >> >> #include<iostream> >> #include<string> >> using namespace std; >> /* >> abasjc a*c >> while(pattern[j] == '*' text[i] == pattern[j]) {i++; j++} >> */ >> bool match(string text, string pattern, int x, int y) >> { >> if(pattern.length() == y) >> { >> cout<<"hey\n"; >> return 1; >> } >> if(text.length() == x) >> { >> cout<<"shit\n"; >> return 0; >> } >> if(pattern[y] == '.' || text[x] == pattern[y]) >> { >> cout<<"in match"<<endl; >> return match(text,pattern,x+1,y+1); >> } >> if(pattern[y] == '*') >> return match(text,pattern,x+1,y) || match(text,pattern,x+1,y+1) >> || match(text,pattern,x,y+1); >> >> if(text[x] != pattern[y]) >> { >> cout<<"shit1\n"; >> return 0; >> } >> >> } >> >> int main() >> { >> string text,pattern; >> cin >> text >> pattern; >> cout << match(text, pattern,0, 0); >> } >> >> On Thu, Dec 27, 2012 at 6:10 PM, shady <sinv...@gmail.com> wrote: >> >>> Thanks for the link Ritesh, >>> if (isMatch(s, p+2)) return true; >>> isnt this line incorrect in the code, as it can lead to segmentation >>> fault... how can we directly access p+2 element, we know for sure that p is >>> not '\0', but p+1 element can be '\0' , therefore leading to p+2 to be >>> undefined. >>> >>> On Thu, Dec 27, 2012 at 6:23 AM, Ritesh Mishra <rforr...@gmail.com>wrote: >>> >>>> try to solve it by recursion .. >>>> http://www.leetcode.com/2011/09/regular-expression-matching.html >>>> >>>> >>>> Regards, >>>> >>>> Ritesh Kumar Mishra >>>> Information Technology >>>> Third Year Undergraduate >>>> MNNIT Allahabad >>>> >>>> >>>> On Sun, Dec 23, 2012 at 11:14 PM, Prem Krishna Chettri < >>>> hprem...@gmail.com> wrote: >>>> >>>>> Well I can tell you Something about design pattern to solve this >>>>> case.. >>>>> >>>>> What I mean is by using The State Machine Design Pattern, >>>>> Anyone can solve this. but Ofcourse it is complicated. >>>>> >>>>> >>>>> >>>>> >>>>> On Sun, Dec 23, 2012 at 11:01 PM, shady <sinv...@gmail.com> wrote: >>>>> >>>>>> that's the point, Have to implement it from scratch... otherwise java >>>>>> has regex and matcher, pattern to solve it........... >>>>>> >>>>>> >>>>>> On Sun, Dec 23, 2012 at 10:28 PM, saurabh singh >>>>>> <saurab...@gmail.com>wrote: >>>>>> >>>>>>> If you need to implement this for some project then python and java >>>>>>> have a very nice library >>>>>>> >>>>>>> >>>>>>> Saurabh Singh >>>>>>> B.Tech (Computer Science) >>>>>>> MNNIT >>>>>>> blog:geekinessthecoolway.blogspot.com >>>>>>> >>>>>>> >>>>>>> On Sun, Dec 23, 2012 at 7:48 PM, shady <sinv...@gmail.com> wrote: >>>>>>> >>>>>>>> >>>>>>>> http://stackoverflow.com/questions/13144590/to-check-if-two-strings-match-with-alphabets-digits-and-special-characters >>>>>>>> >>>>>>>> any solution for this......... we need to implement such regex >>>>>>>> tester................ >>>>>>>> >>>>>>>> some complex cases : >>>>>>>> *string* * regex * -> * status* >>>>>>>> * >>>>>>>> * >>>>>>>> reesd re*.d -> match >>>>>>>> re*eed reeed -> match >>>>>>>> >>>>>>>> can some one help with this ? >>>>>>>> >>>>>>>> -- >>>>>>>> >>>>>>>> >>>>>>>> >>>>>>> >>>>>>> -- >>>>>>> >>>>>>> >>>>>>> >>>>>> >>>>>> -- >>>>>> >>>>>> >>>>>> >>>>> >>>>> -- >>>>> >>>>> >>>>> >>>> >>>> -- >>>> >>>> >>>> >>> >>> -- >>> >>> >>> >> >> >> >> -- >> Cheers, >> >> Vicky >> >> -- >> >> >> > > -- > > > --
