Comments by RastamanVibrations

RastamanVibrations 22-Jan-21 4:39am View

this was what was wrong in the end ,after changing this i no longer get calls for terminations , thanks alot my friend for reading through my code and finding my mistake , i highly apreciate it

RastamanVibrations 22-Jan-21 4:37am View

my friend,thanks again for your previous answer on my previous question , your code although briliantly simple and functional , i cannot use it in a way to provide to the user the steps taken to produce the string . So i had to resort to making another programm that in the end allowed me to change the non-terminal symbols into terminal ones,all this is made randomly using a seeded rand , although i cannot use your code as is , still i highly aprecciate your answer

RastamanVibrations 21-Jan-21 20:09pm View

alright i will see what i can do

RastamanVibrations 21-Jan-21 20:03pm View

so how do i prevent this ?

RastamanVibrations 21-Jan-21 19:53pm View

so the problem i am facing here is that ,by calling the same function over i cause a stack overflow ?

RastamanVibrations 21-Jan-21 19:48pm View

its not a certain line thoe , i get thrown a termination
lets say the next string to be checked again is :
string is (v-Y)
temp = 5
turning E to (Y)--> it means that the search() function(the same you posted in code) detected a letter E while there obviously isnt one
token1 is (v-Y)
terminate called after throwing an instance of 'std::out_of_range'
what(): basic_string::substr: __pos (which is 6) > this->size() (which is 5)

RastamanVibrations 21-Jan-21 19:44pm View

ok, i am sry for the messy code , but as it seems even after changing the sizeof into s.size() i still get the error,how is my search() function detecting an E when the string is (v-Y) or (v+Y) etc.It just doesnt make sense.

RastamanVibrations 21-Jan-21 16:12pm View

hello , and thank you very much for your answer my friend you really helped me out and it works perfectly

Comments by RastamanVibrations (Top 8 by date)