Top Down Parsers - LL(1) Parsers

Compiler Design: Top Down Parsers - LL(1) Parsers
Topics discussed:
1. Understanding the significance of the name of Recursive Descent Parsers.
2. Organization of the LL(1) Parsers.
3. Why LL(1) Parsers are called LL(1) Parsers?
4. Understanding the concepts of FIRST() and FOLLOW() with examples.
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Axol x Alex Skrindo - You [NCS Release]
#CompilerDesignByNeso #CompilerDesign #Parsers #LL1Parsers

[@doniakh7482]

this has helped me more than my 2 hours lecture... thank you so much

[@gatecomputerscience1484]

so beautifully explain

[@kajalsharma5952]

so beautiful , so elegant just looking like a wow! amazing series .......please complete it before jan my end sems a whole hearted request! 🥺❣

[@siddireddyvignesh]

Already completed

[@nikhilmaid2670]

are u gonna complete the aptitude and reasoning playlist??

[@dinushachathuranga7657]

Thanks for the nice explanation💜

[@reyy9220]

dude thank you soooo much ur a lifesaver

[@509nandini7]

Sir here we no need find the FIRST (A) FIRST (B) or We have to do that...?

[@omarzakzook6077]

why in the first(S) w didn't put c ??

[@sahilyadav3782]

because we can't have a c in the first place when S is substituted the grammar was S -> ABC A -> a | e (think of e as epsilon please) B -> b C -> c now, S is substituted by A A is then substituted by a, which is a terminal So, we can have an `a` at the first position when S is substituted with terminals Hence we put `a` in the set of FIRST(S) i.e. FIRST(S) = {a} Now, A can also be substituted by e(epsilon) In this case, when we move ahead with S -> ABC, after A comes B, and B->b (from the grammar) Since epsilon means null and is not written in the final string of terminals, in this specific case, when S is substituted, it starts with a `b` That means, S can also be substituted by a terminal string which starts with `b`. So we can push `b` in the FIRST(S) FIRST(S) = {a, b} But in no case can there be a terminal string for S which starts with C, cause B will always derive b, and since B comes before C in S, i.e. S -> ABC, S will always have at least a b before c. Hence, there will be no terminal string for S which starts with c, so we don't put c in the FIRST(S)