Code Optimization Questions and Answers

Q: State whether the following is true or false.A compiler may freely reorder floating point additions to expose parallelism.

Correct answer: FalseReordering additions changes where the rounding occurs and so can change the answer, sometimes substantially in an ill-conditioned sum. A conforming compiler will not do it by default, which is precisely why flags exist to permit it - and why a program that suddenly gives different results under such a flag is behaving as documented rather than as a defect.

Q: A program whose behaviour changes when optimisation is enabled most often indicates

Correct answer: undefined behaviour in the program itselfAn optimiser is entitled to assume the program does not do anything the language leaves undefined - no signed overflow, no reading uninitialised storage, no aliasing violations. A program relying on what happened to work unoptimised has usually been depending on one of those, and the transformation has merely exposed it.

51.

An optimisation that is legal for integer arithmetic may be illegal for floating point because

A.the compiler cannot determine floating point types
B.floating point arithmetic is always slower to perform
C.floating point operations are not associative
D.floating point values cannot be held in registers

Answer: C

Correct answer: floating point operations are not associative

\( (a+b)+c \) and \( a+(b+c) \) can differ in floating point because each addition rounds, so reassociating changes the result. Integer arithmetic is exactly associative, so the same transformation is safe there. This is why compilers require an explicit fast-math flag before reassociating floating point.

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52.

State whether the following is true or false.
A compiler may freely reorder floating point additions to expose parallelism.

A.True only at large scales
B.False only at small scales
C.True
D.False

Answer: D

Correct answer: False

Reordering additions changes where the rounding occurs and so can change the answer, sometimes substantially in an ill-conditioned sum. A conforming compiler will not do it by default, which is precisely why flags exist to permit it - and why a program that suddenly gives different results under such a flag is behaving as documented rather than as a defect.

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53.

An optimisation must be suppressed for a variable declared volatile because

A.its value may change by means the compiler cannot see
B.volatile variables cannot be held in machine registers
C.the language standard forbids optimising such programs
D.volatile variables always occupy more storage than others

Answer: A

Correct answer: its value may change by means the compiler cannot see

A volatile object may be a memory-mapped device register or a location written by another thread or a signal handler, so the compiler’s reasoning about what the program does to it is incomplete. Every read and write must therefore be performed exactly as written, in order, and none may be cached in a register or elided.

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54.

Aliasing limits optimisation because

A.aliases always indicate an error in the source program
B.two names may refer to the same location, so a write to one affects both
C.aliased variables cannot be stored in memory at all
D.the compiler is quite unable to determine the types of the aliased names

Answer: B

If the compiler cannot prove two pointers are distinct it must assume they may coincide. That assumption blocks reordering, caching in registers and much else. Languages with restricted pointers give the optimiser far more freedom.

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55.

Alias analysis is described as conservative when it

A.ignores pointers entirely and considers only named variables
B.assumes that no two pointers ever refer to the same object
C.reports a possible alias whenever it cannot prove there is none
D.reports an alias only when it can prove one exists

Answer: C

Assuming distinctness without proof would produce wrong code. Reporting a possible alias costs an optimisation, which is the acceptable side on which to err. Type based rules let compilers rule out some aliases without any analysis.

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56.

The phase ordering problem in optimisation refers to

A.the compiler running its phases in the wrong sequence
B.the sheer impossibility of parsing before the source has been scanned
C.the need to allocate registers before generating code
D.one transformation creating or destroying opportunities for another

Answer: D

Correct answer: one transformation creating or destroying opportunities for another

Inlining exposes constants that propagation can use; propagation exposes dead code; dead code elimination may make a loop invariant. But an ordering good for one program may be poor for another, and no single sequence is best in general - which is why compilers run some passes repeatedly rather than once each.

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57.

Assertion (A): Optimisation passes are often run more than once in a compiler.
Reason (R): One transformation frequently exposes opportunities that a pass which already ran would have taken.

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

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. Because transformations expose opportunities for one another, a pass that ran early may have missed work that only becomes visible later. Running the pipeline round again is the practical response to a phase ordering problem that has no optimal solution.

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58.

The optimisation level selected by a compiler flag typically controls

A.the number of errors the compiler is willing to report
B.which transformations run and how much compilation time is spent
C.whether the program will be correct at all when it comes to be executed
D.the source language that the compiler will accept

Answer: B

Higher levels apply more and costlier analyses. Correctness is not negotiable at any level, though a bug in an optimisation may of course surface only at the higher settings. That is why a program failing only when optimised deserves careful examination.

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59.

A program compiled without optimisation differs from one compiled with it in that the unoptimised program

A.cannot be debugged with a symbolic debugger
B.contains fewer instructions in the generated code
C.runs more slowly but computes the same results
D.may produce different results from the optimised version

Answer: C

Optimisation is meant to preserve behaviour exactly. A difference in results indicates either a compiler defect or a program relying on behaviour the language leaves undefined. The latter explanation is by far the more common of the two.

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60.

A program whose behaviour changes when optimisation is enabled most often indicates

A.a defect in the optimiser that must be reported
B.that the program was compiled for the wrong machine
C.that the source language has been misunderstood
D.undefined behaviour in the program itself

Answer: D

Correct answer: undefined behaviour in the program itself

An optimiser is entitled to assume the program does not do anything the language leaves undefined - no signed overflow, no reading uninitialised storage, no aliasing violations. A program relying on what happened to work unoptimised has usually been depending on one of those, and the transformation has merely exposed it.

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