• ( 1 ) Uniform symbol table

    • 1) contains all constants in the program
    • 2) consists of full or partial list of the token's as they appear in the program created by Lexical analysis and used for syntax analysis and interpretation.
    • 3) a permanent table which lists all key words and special symbols of the language in symbolic form
    • 4) None of these
    • Discussion in forum
      Answer : 3) a permanent table which lists all key words and special symbols of the language in symbolic form
      Solution :








      discussion


      Answer : 3) a permanent table which lists all key words and special symbols of the language in symbolic form

    • ( 2 ) The part of the machine level instruction, which tells the central processor what has to be done is

    • 1) Operation code
    • 2) Address
    • 3) Locator
    • 4) Flip Flop
    • Discussion in forum
      Answer : 1) Operation code
      Solution :








      discussion


      Answer : 1) Operation code

    • ( 3 ) Which table is permanent databases that has an entry for each terminal symbol ?

    • 1) Terminal table
    • 2) Literal table
    • 3) Literal table
    • 4) None of these
    • Discussion in forum
      Answer : 1) Terminal table
      Solution :








      discussion


      Answer : 1) Terminal table

    • ( 4 ) A compiler for a high-level language that runs on one machine and produces code for a different machine is called

    • 1) optimizing compiler
    • 2) one pass compiler
    • 3) cross compiler
    • 4) multipass compiler
    • Discussion in forum
      Answer : 3) cross compiler
      Solution :








      discussion


      Answer : 3) cross compiler

    • ( 5 ) Peep-hole optimization is a form of

    • 1) loop optimization
    • 2) local optimization
    • 3) constant folding
    • 4) data flow analysis
    • Discussion in forum
      Answer : 3) constant folding
      Solution : Redundant instructions may be discarded during the final stage of compilation by using a simple optimizing technique called peephole optimization.It is a kind of optimization performed over a very small set of instructions in a segment of generated code. The set is called a "peephole" or a "window". It works by recognising sets of instructions that can be replaced by shorter or faster sets of instructions and it uses some common techniques : Constant folding. So option (3) is correct








      discussion


      Answer : 3) constant folding

    • ( 6 ) The table created by lexical analysis to describe all literals used in the source program is

    • 1) Terminal table
    • 2) Literal table
    • 3) Identiier table
    • 4) Reductions
    • Discussion in forum
      Answer : 2) Literal table
      Solution :








      discussion


      Answer : 2) Literal table

    • ( 7 ) Which of the following features cannot be captured by CFG ?

    • 1) syntax of if-then-else statements
    • 2) syntax of recursive procedures
    • 3) whether a variable is declared before its use
    • 4) matching nested paranthesis
    • Discussion in forum
      Answer : 4) matching nested paranthesis
      Solution : it is because, it is equivalent to recognizing wew, where the first w is the decleration and the second is its use, wew is not a CFG.








      discussion


      Answer : 4) matching nested paranthesis

    • ( 8 ) The method which merges the bodies of two loops is

    • 1) loop rolling
    • 2) loop Jamming
    • 3) constant folding
    • 4) None of these
    • Discussion in forum
      Answer : 2) loop Jamming
      Solution :








      discussion


      Answer : 2) loop Jamming

    • ( 9 ) Consider the grammar:
      S → (S) | a
      Let the number of states in SLR(1), LR(1) and LALR(1) parsers for the grammar be n1, n2 and n3 respectively. The following relationship holds good

    • 1) n1 < n2 < n3
    • 2) n1 = n3 < n2
    • 3) n1 = n2 = n3
    • 4) n1 ≥ n3 ≥ n2
    • Discussion in forum
      Answer : 2) n1 = n3 < n2
      Solution : LALR(1) is formed by merging states of LR(1) ( also called CLR(1)),
      hence no of states in LALR(1) is less than no of states in LR(1),
      therefore n3 < n2.
      And SLR(1) and LALR(1) have same no of states, i.e ( n1 = n3).
      Hence n1 = n3 < n2








      discussion


      Answer : 2) n1 = n3 < n2

    • ( 10 ) Which one of the following is True at any valid state in shift-reduce parsing?

    • 1) Viable prefixes appear only at the bottom of the stack and not inside
    • 2) Viable prefixes appear only at the top of the stack and not inside
    • 3) The stack contains only a set of viable prefixes
    • 4) The stack never contains viable prefixes
    • Discussion in forum
      Answer : 3) The stack contains only a set of viable prefixes
      Solution :








      discussion


      Answer : 3) The stack contains only a set of viable prefixes





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