Intermediate Code Generation Questions and Answers

Q: State whether the following is true or false.Every value computed during expression translation must be stored in memory.

Correct answer: FalseMost intermediate values are held in temporaries which the register allocator will place in registers, and never reach memory at all. Storing every value would make the generated code far slower than it needs to be, and avoiding that is a large part of what code generation is for.

51.

Arrange these steps of partitioning three address code into basic blocks.
1. mark each jump target and each instruction after a jump
2. identify the leaders among the instructions
3. form a block from each leader up to the next leader
4. take the first instruction of the program as a leader

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

Answer: C

The leaders are found first by the two rules, and each block then runs from one leader to just before the next. The partition is unique for a given instruction sequence. Two compilers applying the rules will always agree on the blocks.

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52.

An instruction is a leader of a basic block if it

A.assigns a value to a compiler generated temporary
B.contains an operator with two operands rather than one
C.appears at an even numbered position in the sequence
D.is the first instruction, a jump target, or follows a jump

Answer: D

Those three rules identify every point at which control may enter. Applying them to a sequence of instructions yields the partition into blocks directly. No judgement is involved, so the partition can be computed mechanically.

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53.

A three address program of 20 instructions has 4 leaders. How many basic blocks does it contain?

A.4
B.5
C.16
D.20

Answer: A

A leader is the first instruction of a basic block, and every block has exactly one leader, so the two counts are equal: \[ \text{blocks} = \text{leaders} = 4 \] The 20 instructions are partitioned among those four blocks, with none left over and none belonging to two blocks at once. The instruction count constrains the average block size, \( 20/4 = 5 \), but not the number of blocks.

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54.

A flow graph is constructed from basic blocks by

A.merging all the blocks that contain no jump instructions
B.adding an edge wherever control can pass from one block to another
C.connecting every block to every other block in the program
D.ordering the blocks by the number of instructions each contains

Answer: B

Correct answer: adding an edge wherever control can pass from one block to another

The nodes are the basic blocks and the edges are the possible transfers between them - a jump, a conditional branch taken or not taken, or simply falling through. That graph is what every data flow analysis propagates its facts over, so building it correctly is a precondition for optimising at all.

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55.

The intermediate representation is a good place to perform optimisation because

A.the parser has not yet finished processing the program
B.no temporaries have been introduced at that stage
C.it is independent of both the source language and the machine
D.it contains fewer instructions than the target code will

Answer: C

A transformation written once at that level benefits every source language and every target. Machine specific optimisations are left for the back end where the instruction set is known. The division keeps the front end usable with any target.

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56.

Translating a nested expression requires temporaries because

A.the target machine has no instructions for nested expressions
B.the parser cannot recognise nesting deeper than three levels
C.the symbol table has no room for the intermediate values
D.a three address instruction cannot name a subexpression directly

Answer: D

Each instruction names addresses, not expressions, so every intermediate value needs somewhere to live. The number of temporaries equals the number of operators under the simple scheme. Reusing them where lifetimes do not overlap reduces that count considerably.

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57.

Reusing temporary names where their lifetimes do not overlap

A.reduces the number of distinct names the back end must handle
B.changes the meaning of the program being compiled
C.is impossible once the intermediate code has been generated
D.requires the program to be converted to single assignment form

Answer: A

Correct answer: reduces the number of distinct names the back end must handle

Expression translation generates a great many temporaries, most of them short-lived, and reusing a name once its value is dead cuts the number the register allocator must consider. The program’s meaning is unaffected, since the values never coexist.

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58.

State whether the following is true or false.
Every value computed during expression translation must be stored in memory.

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

Answer: B

Correct answer: False

Most intermediate values are held in temporaries which the register allocator will place in registers, and never reach memory at all. Storing every value would make the generated code far slower than it needs to be, and avoiding that is a large part of what code generation is for.

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59.

A copy instruction of the form x = y in intermediate code

A.is required by the language standard after every assignment
B.indicates that the two names refer to different types
C.is often removable by a later copy propagation pass
D.cannot be removed without changing the program's meaning

Answer: C

Correct answer: is often removable by a later copy propagation pass

Copy propagation replaces later uses of x with y, after which the copy itself may be dead and removable. Copies arise in quantity from the translation of assignments and from parameter passing, so this is one of the more productive of the simple optimisations.

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60.

Translating a pointer dereference produces an instruction of the form

A.x = &y, taking the address of the named variable
B.x = y[i], indexing an array by a computed subscript
C.param y, which sets up one argument for a procedure call
D.x = *y, an indirect assignment reading through the pointer

Answer: D

Correct answer: x = *y, an indirect assignment reading through the pointer

Three address code needs distinct forms for the pointer operations: taking an address, reading through a pointer, and writing through one. The dereference reads the value at the address held in y. Its presence complicates optimisation greatly, since the compiler may not know what y points at.

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Intermediate Code Generation Questions and Answers

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