Code Optimization Questions and Answers

Q: Local optimisation is optimisation performed

Correct answer: within a single basic block at a timeA basic block is a straight-line run with one entry and one exit, so within it control flow is trivial and every statement executes exactly once in order. That makes local optimisation cheap and certain - no data flow analysis is needed - and correspondingly limited in what it can see.

Q: Match each scope of optimisation with the region it considers.List IA. localB. globalC. interproceduralD. peepholeList II1. several procedures together2. one basic block3. a short window of adjacent instructions4. one whole procedure

Correct answer: A-2, B-4, C-1, D-3Four scopes in increasing order of reach and cost. Peephole works over a short window of adjacent instructions; local optimisation over one basic block; global optimisation over a whole procedure, which is where data flow analysis becomes necessary; and interprocedural optimisation across several procedures at once.

The overriding requirement on any code optimisation is that it must

A.preserve the meaning of the program for every possible input
B.reduce the number of instructions in the generated code
C.shorten the time the compiler itself takes to run
D.be applicable to every program the compiler will ever see

Answer: A

Correct answer: preserve the meaning of the program for every possible input

Correctness is the constraint within which everything else is negotiable. A transformation that makes the program faster on the cases tested and wrong on one it was not is not an optimisation at all, and this is why compilers reason conservatively and decline transformations they cannot prove safe.

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Local optimisation is optimisation performed

A.on the source text before parsing has begun
B.within a single basic block at a time
C.across the whole of a single procedure at once
D.across every procedure in the whole program

Answer: B

Correct answer: within a single basic block at a time

A basic block is a straight-line run with one entry and one exit, so within it control flow is trivial and every statement executes exactly once in order. That makes local optimisation cheap and certain - no data flow analysis is needed - and correspondingly limited in what it can see.

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Match each scope of optimisation with the region it considers.
List I
A. local
B. global
C. interprocedural
D. peephole
List II
1. several procedures together
2. one basic block
3. a short window of adjacent instructions
4. one whole procedure

A.A-2, B-4, C-3, D-1
B.A-3, B-1, C-4, D-2
C.A-2, B-4, C-1, D-3
D.A-4, B-2, C-1, D-3

Answer: C

Correct answer: A-2, B-4, C-1, D-3

Four scopes in increasing order of reach and cost. Peephole works over a short window of adjacent instructions; local optimisation over one basic block; global optimisation over a whole procedure, which is where data flow analysis becomes necessary; and interprocedural optimisation across several procedures at once.

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Splitting a procedure into basic blocks is worth doing before optimisation because

A.each block can then be compiled by a separate pass of the compiler
B.the blocks can be reordered freely without affecting the program
C.a block always corresponds to a single statement of the source
D.reasoning within a block needs no knowledge of control flow at all

Answer: D

Since control enters a block only at the top and leaves only at the bottom, every instruction runs whenever the block does. That guarantee is what makes local transformations safe without any data flow analysis, and it is why the partition is computed before anything else.

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The partition of a program into basic blocks is

A.unique, since the leader rules leave no room for choice
B.one of many possible partitions the compiler may select
C.chosen so that every block has the same instruction count
D.determined by the source syntax rather than by the code

Answer: A

The three leader rules are entirely mechanical, so any two compilers applying them to the same instruction sequence produce exactly the same blocks. Nothing about the source syntax or the block sizes enters into the computation at all.

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A flow graph has as its nodes

A.the variables that the program declares and uses
B.the basic blocks of the procedure being analysed
C.the individual instructions of the whole program
D.the procedures declared in the compilation unit

Answer: B

Correct answer: the basic blocks of the procedure being analysed

The flow graph abstracts away everything inside a block and keeps only the possible transfers of control between blocks. This is exactly the right level for data flow analysis, which asks what holds on entry to and exit from each block and propagates those facts along the edges.

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Constant folding is the transformation that

A.removes a statement whose result is never used anywhere
B.moves a computation right out of a loop that does not ever change it
C.works out at compile time the value of an expression built from literals
D.replaces a variable by the constant it is known to hold

Answer: C

Correct answer: works out at compile time the value of an expression built from literals

Constant folding evaluates \( 3 \times 4 \) to 12 during compilation. It is distinct from constant propagation, which replaces a variable by a constant known to be in it - the two are almost always run together, each exposing further opportunities for the other.

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Constant propagation is the transformation that

A.evaluates an expression whose operands are all literal constants
B.eliminates a variable that is assigned but never read
C.reorders instructions so constants are computed first
D.replaces a use of a variable by the constant it must hold there

Answer: D

Establishing that only one constant reaches the use is a data flow question. Propagation and folding reinforce one another, each creating opportunities for the other. Running the pair repeatedly until nothing changes is common practice.

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Common subexpression elimination replaces a repeated computation provided that

A.none of its operands has been changed since the first evaluation
B.the expression turns up at least three times somewhere in the procedure
C.all the operands of the expression are compile time constants
D.there are registers to spare on whatever machine is being targeted

Answer: A

An assignment to any operand between the two occurrences invalidates the saved value. Establishing that no such assignment occurs requires available expression analysis. Within a single block the check is much simpler and needs no data flow at all.

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10.

Dead code elimination is usually run more than once during compilation because

A.the pass must be repeated once for every basic block present
B.earlier transformations keep making further code dead
C.the pass is unable to remove all the dead code in one go
D.each run removes only the first dead statement it encounters

Answer: B

Constant propagation may make a branch unreachable and copy propagation may leave an assignment with no remaining uses. Each round of removal can therefore expose more dead code, so the pass is iterated until nothing further changes.

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