Accumulators in PROLOG

[@marcosalmistraro7869]

this whole concept is just infuriating. Following a course on Prolog right now and really can't get my head around it. Thanks for trying to explain !

[@edwardgrabczewski]

I find recursive programming rather difficult to follow and you do this video at a fast speed. All my experience of programming is with procedural languages like C and PL/SQL. I can see now why Lisp, Prolog and Logo didn't become popular for general programming since recursion, as opposed to iteration, is quite mind-bending! Does recursion get any easier with practice?

[@techdose4u]

Recursion gets pretty easier with practice. I have made a separate video for Backtraking in PROLOG. Do check it out to get a better understanding of PROLOG Recursion.

[@estetlinus1]

I totally agree. This is as non-intuitive as it gets :'-(

[@bal3a100]

Great Work :D

[@techdose4u]

Thnq for inspiring :) pls share the video to those who needs it.

[@bal3a100]

@@techdose4u Yes I will do, please share more videos we need them asap

[@techdose4u]

@@bal3a100 Do you need backtracking and cut?? or do you need to understand some code? COMMENT your preferences and accordingly i will make them ASAP

[@bal3a100]

@@techdose4u about append, some sorts ( merge ; quicksort, ..) ; backtracking and cut That will be awesome THX :D

[@shresthmishra9329]

If anyone wants to execute it-> listlen(L, N) :- listacc(L, 0, N). listacc([], A, A). listacc([_|T], A, N) :- A1 is A+1, listacc(T, A1, N).

[@techdose4u]

🙏

[@shresthmishra9329]

@@techdose4u sir cse to le liya tha bina soche samghe lekin sab paar apk tutorials se hi ho rha hai🙏🙏🙏❤️❤️❤️

[@techdose4u]

Ho jaega chinta mt kro :) Bas lagey rho.

[@Huda-dq9tm]

I tried running both the codes and both showed a same error that said "Singleton Variable: H". I do not understand the error but it went away when I replaced H with an underscore and then the programs worked satisfactorily for all kinds of test cases. Can you explain why this happened?

[@YoungVinny]

Because H is only being used once so Prolog doesn't like that and wants you to use an anonymous variable (underscore)

[@shresthmishra9329]

listlen(L, N) :- listacc(L, 0, N). listacc([], A, A). listacc([_|T], A, N) :- A1 is A+1, listacc(T, A1, N). Try this . It will work.

[@Huda-dq9tm]

Shresth that's so nice of you but honestly it was so long ago. I dont even remember why I asked what I asked. But thankyou!

[@videosforyou567]

@@Huda-dq9tm @Shresth Mishra Do you guys use prolog now?

[@alaukiktandlekar114]

I have typed code as shown but i am getting output 0 why?

[@MahmoudEssamMustafa]

listLen(List,N):- lenAcc(List,0,N). lenAcc([],A,A). lenAcc([_|Tail],A,N):- A1 is A+1,lenAcc(Tail,A1,N).

[@MahmoudEssamMustafa]

in line 3 you should use the same predicate name as in line 2.

[@muyybe]

unclear explanation and no advice for tgese problems.