Intermediate Code Generation Questions and Answers

Q: Arrange the steps of translating an assignment statement into three address code.1. generate the instruction combining the operand temporaries2. parse the right hand side into a tree or during the parse3. generate the copy assigning the result to the left hand side4. generate code for each subexpression into a temporary

Correct answer: 2 - 4 - 1 - 3The order follows the structure of the expression. The right hand side is parsed, each subexpression is evaluated into a temporary, those temporaries are combined by the operator, and the result is finally copied to the left hand side. Generating the copy first would have nothing to copy.

11.

Assertion (A): Quadruples are preferred to triples when the intermediate code will be optimised.
Reason (R): A triple names its result by its own position, so moving an instruction invalidates every reference to it.

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

Answer: C

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. Optimisation moves and removes instructions constantly, and in a triple representation an instruction’s identity is its position, so every such change invalidates the references to it. A quadruple names its result explicitly and is unaffected.

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12.

Indirect triples represent a program as

A.triples in which every operand is itself a triple
B.quadruples with the result field left permanently empty
C.three separate arrays holding operators and operands
D.a list of pointers into an array of triples

Answer: D

Reordering then means permuting the pointer list while the triples themselves stay put. This restores the reorderability of quadruples while still sharing identical subexpressions. The extra indirection costs one memory reference per access.

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13.

Match each representation with its distinguishing property.
List I
A. quadruple
B. triple
C. indirect triple
D. static single assignment
List II
1. a pointer list over a triple array
2. explicit result field
3. each name assigned exactly once
4. result named by instruction position

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

Answer: A

Each form trades space against ease of manipulation. Choosing among them is essentially a decision about how much optimisation the compiler intends to perform on the representation. A compiler that optimises little can afford the more compact forms.

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14.

In static single assignment form

A.no temporaries are introduced during the translation
B.every variable is assigned a value at exactly one point
C.every instruction has exactly one operand and one result
D.the program contains exactly one basic block throughout

Answer: B

Renaming each assignment to a fresh version makes the definition reaching any use immediately apparent. Many data flow analyses become far simpler and faster on this form. Modern optimising compilers use it as their main internal representation.

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15.

A phi function in static single assignment form is needed

A.at the entry to every procedure in the program
B.whenever an expression contains more than two operands
C.where control flow paths carrying different versions of a name meet
D.wherever a variable happens to be assigned more than once in a block

Answer: C

Correct answer: where control flow paths carrying different versions of a name meet

Static single assignment gives every assignment a distinct name, which works within straight-line code but fails at a join, where the value could have come from either path. The phi function is placed at that join to select between the incoming versions, which restores the single-assignment property.

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16.

A syntax tree used as an intermediate representation differs from three address code in that it

A.contains strictly fewer nodes than the equivalent code
B.is unable to represent an expression with more than two operands
C.records the machine registers assigned to each subexpression
D.retains the nesting structure rather than flattening it

Answer: D

The tree is closer to the source and easier to transform structurally. Three address code is closer to the machine and easier to schedule and allocate registers for. Most compilers keep both forms and lower from one to the other.

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17.

A directed acyclic graph used as an intermediate representation differs from a syntax tree by

A.sharing a single node between identical subexpressions
B.removing all the operator nodes from the structure
C.allowing cycles that represent loops in the program
D.recording the order in which operations must execute

Answer: A

Common subexpressions are identified by construction, since a node is reused rather than duplicated. This makes the graph a natural vehicle for local common subexpression elimination. The sharing happens as a side effect of building the structure at all.

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18.

The expression a + b + a + b is represented as a directed acyclic graph with sharing. How many distinct interior nodes does it have?

Answer: B

In a directed acyclic graph, identical subexpressions are built once and shared. Here \( a+b \) occurs twice but becomes a single node, and the outer addition that combines the two references is a second: \[ \text{interior nodes} = 1 \ (a+b) + 1 \ (\text{outer} +) = 2 \] A plain syntax tree, which cannot share, would carry three interior nodes for the same expression.

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19.

Constructing a directed acyclic graph for a basic block reveals

A.the syntax errors that the parser failed to detect
B.the number of passes the compiler will need to make
C.common subexpressions and values that are never used
D.the registers that the target machine has available

Answer: C

A node with no parent computes a value nobody consumes, and a shared node is a repeated computation. Both observations feed directly into local optimisation of the block. The graph is therefore built as much for analysis as for representation.

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20.

Arrange the steps of translating an assignment statement into three address code.
1. generate the instruction combining the operand temporaries
2. parse the right hand side into a tree or during the parse
3. generate the copy assigning the result to the left hand side
4. generate code for each subexpression into a temporary

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

Answer: D

Correct answer: 2 - 4 - 1 - 3

The order follows the structure of the expression. The right hand side is parsed, each subexpression is evaluated into a temporary, those temporaries are combined by the operator, and the result is finally copied to the left hand side. Generating the copy first would have nothing to copy.

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Intermediate Code Generation Questions and Answers

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