Intermediate Code Generation Questions and Answers

Q: The number of three address instructions generated for a program is typically

Correct answer: larger than the number of source statementsOne source statement expands into several three address instructions, because each operator and each subexpression becomes an instruction of its own with a temporary to hold the result. An expansion of several times is normal, and it is why the intermediate form is verbose but simple.

61.

Assertion (A): Pointer assignments make optimisation of intermediate code harder.
Reason (R): The compiler cannot always determine which variables a pointer might refer to.

A.Both A and R are true, and R is the correct explanation of A
B.Both A and R are true, but R is not the correct explanation of A
C.A is true but R is false
D.A is false but R is true

Answer: A

Both are true and the reason gives the mechanism, and the difficulty has a name. Alias analysis attempts to bound the set of possible targets, and where it cannot, the optimiser must assume the worst and forgo the transformation.

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62.

Alias analysis attempts to determine

A.the widths of the types declared in the program
B.which names may refer to the same storage location
C.which variables are live at each point in the program
D.the number of times a loop will execute at run time

Answer: B

Two names that may alias cannot be treated independently. The analysis is conservative, reporting a possible alias whenever it cannot prove the absence of one. An unproven possibility must be treated exactly as a certainty would be.

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63.

Generating intermediate code for a nested procedure requires recording

A.the registers the procedure will use at run time
B.the source line on which the procedure was declared
C.the nesting depth so that non local names can be reached
D.the number of instructions the procedure will occupy

Answer: C

Access to a variable in an enclosing procedure needs the depth difference, which selects how many links to follow. Access links or a display implement the traversal at run time. Languages without nested procedures avoid the whole mechanism.

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64.

The intermediate code for a return statement typically

A.deallocates the storage of every variable in the program
B.jumps to the first instruction of the calling procedure
C.requires no instructions at all
D.moves the result to an agreed place and transfers control back

Answer: D

The calling convention fixes where the result is left. The actual return sequence, including any stack adjustment, is completed by the code generator using target conventions. The intermediate form deliberately leaves those details unspecified.

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65.

A compiler that generates intermediate code for a construct it cannot optimise

A.still produces correct if slower target code for it
B.must reject the program as unable to be compiled
C.generates no code at all for that construct
D.converts the construct into a procedure call instead

Answer: A

Optimisation is always optional. Correct translation is the requirement, and an unoptimised construct simply runs more slowly than it might otherwise have done. Correctness is never negotiable in the way performance is.

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66.

The number of three address instructions generated for a program is typically

A.independent of the size of the source program
B.larger than the number of source statements
C.exactly equal to the number of source statements
D.smaller than the number of source statements

Answer: B

Correct answer: larger than the number of source statements

One source statement expands into several three address instructions, because each operator and each subexpression becomes an instruction of its own with a temporary to hold the result. An expansion of several times is normal, and it is why the intermediate form is verbose but simple.

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67.

A compiler using a directed acyclic graph for each basic block can perform

A.instruction scheduling for the target pipeline
B.interprocedural analysis across compilation units
C.local common subexpression elimination during construction
D.global register allocation across the whole procedure

Answer: C

Sharing a node instead of creating a duplicate is exactly the elimination. The other three transformations need information beyond the single block the graph describes. A flow graph over the whole procedure is required for those.

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68.

Value numbering assigns the same number to two expressions when

A.they appear at the same position in two basic blocks
B.they use the same operator regardless of operands
C.they were written on the same line of the source
D.they are known to compute the same value

Answer: D

Equal numbers mark redundant computations that can be shared. The technique is the algorithmic counterpart of building a directed acyclic graph for the block. Both methods find the same redundancies by different routes.

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69.

Which of the following is NOT a common intermediate representation?

A.the token stream produced by the lexical analyser
B.three address code
C.static single assignment form
D.a directed acyclic graph built over one basic block

Answer: A

The first three are all representations of the program's computation. The token stream carries no structure at all and is consumed by the parser rather than passed to the back end. By the time code is generated the tokens have long since been discarded.

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70.

Intermediate code generation is placed after semantic analysis because

A.optimisation must be performed before types are checked
B.the types of expressions must be known before code can be emitted
C.the parser cannot complete until code has been generated
D.the symbol table is not created until the code generation begins

Answer: B

Which instruction to emit depends on the operand types, and any coercions must already have been inserted. Generating code first would mean regenerating it once the types became known. Coercions in particular change which instruction must be emitted.

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