Skip to content

Compiler Design CAE 3 Question Bank

Answers

1. How will you group the phases of compiler

The phases of a compiler are typically grouped into two main categories: the frontend and the backend. Each of these categories encompasses several phases that work together to translate source code into machine code or an intermediate representation. Here's how the phases of a compiler are commonly grouped:

  1. Frontend: The frontend of a compiler is responsible for the initial processing of the source code, including syntax analysis and semantic analysis. It ensures that the source code is well-formed and valid, preparing it for further translation and optimization. The frontend phases include:

    a. Lexical Analysis (Scanner):

    • This phase breaks the source code into tokens or lexemes.
    • It removes comments, whitespace, and irrelevant details from the source code.
    • It identifies keywords, identifiers, constants, and operators.

    b. Syntax Analysis (Parser):

    • This phase checks the syntax of the source code to ensure it adheres to the language's grammar rules.
    • It builds a parse tree or abstract syntax tree (AST) that represents the structure of the code.

    c. Semantic Analysis:

    • This phase checks the meaning and context of the code.
    • It enforces language-specific rules, type checking, and scope resolution.
    • It may generate a symbol table to store information about identifiers.

    d. Intermediate Code Generation:

    • Some compilers generate an intermediate representation of the source code for further optimization and translation.
    • Intermediate code may be in the form of three-address code, bytecode, or an abstract intermediate language.

    e. Optimization (Optional):

    • This phase may include various optimization techniques to improve the efficiency of the generated code.
    • Common optimizations include constant folding, common subexpression elimination, and dead code elimination.

  2. Backend: The backend of a compiler focuses on code generation and optimization to produce the target machine code or intermediate representation. The backend phases include:

    a. Intermediate Code Translation (if applicable):

    • If an intermediate representation was generated during the frontend, it is translated into target machine code or another intermediate form that is closer to the target architecture.

    b. Code Optimization:

    • The compiler performs various optimizations to improve the efficiency of the generated code.
    • This includes optimization techniques such as register allocation, loop optimization, and instruction scheduling.

    c. Code Generation:

    • In this phase, the compiler generates the final target machine code or executable program.
    • It maps the high-level language constructs to the corresponding machine code instructions.

    d. Object File Generation:

    • The compiler may produce object files or binary files that can be linked together to create the final executable program.

    e. Linking (if applicable):

    • Linker and loader processes may be involved in combining multiple object files and libraries to produce the final executable program.

2. What are the various parts in LEX program

In a Lex program (often referred to as a lexical analyzer or lexer), there are several key parts or components that work together to analyze and tokenize input text. Lex is a tool for generating lexical analyzers based on regular expressions. Here are the main parts of a Lex program:

  1. Lexical Rules:

    • Lex programs start with a set of regular expressions and corresponding actions that define the lexical rules. These rules specify patterns to be matched in the input text.

    lex

  2. %% [0-9]+ { printf("Found an integer: %s\n", yytext); } [a-zA-Z]+ { printf("Found an identifier: %s\n", yytext); } . { printf("Found an unknown character: %c\n", *yytext); } %%

    In this example, the lexical rules define how to recognize integers, identifiers, and other characters.

  3. User-Defined Actions:

    • User-defined actions are associated with each regular expression and are executed when a match is found. These actions define what should be done when a pattern is recognized.

    • Lexical Analyzer Initialization:

    • The Lex program typically includes a section for initialization, which can include variable declarations and other setup code.

    lex

  4. %% int line_number = 1; %%

    In this example, a variable line_number is initialized to 1.

  5. Lexical Analyzer Code:

    • The core of the Lex program is the code that reads and analyzes the input text. This code is often automatically generated by Lex based on the provided rules and actions.

    • Regular Expressions and Actions Separator:

    • In a Lex program, the double percentage symbols (%%) are used to separate the lexical rules and actions from the initialization and any other code.

3. Define left recursion. Is the following grammar left recursive?

E → E+E | E*E| a| b

Left recursion is a grammatical structure in which a non-terminal symbol can derive itself directly or indirectly through a series of productions. In other words, a grammar is left-recursive if there is a production where the non-terminal symbol appears as the leftmost symbol on the right-hand side. Left-recursive grammars can be problematic for some parsing algorithms, so they are often avoided.

Let's examine the given grammar:

E → E + E | E * E | a | b

In this grammar, there are two productions that are potentially left-recursive:

  1. E → E + E
  2. E → E * E

Both of these productions have E as the leftmost symbol on the right-hand side. This indicates left recursion.

To eliminate left recursion from a grammar, you can use left-factoring or rewrite the grammar in a non-left-recursive form. Here's the modified grammar that eliminates left recursion:

E → T E' E' → + T E' | * T E' | ε T → a | b

In this modified grammar:

  • E derives T followed by E', where T represents a single term (either a or b).
  • E' allows for zero or more additions or multiplications, using the + and * operators, followed by another term and another E'.
  • The ε (epsilon) in the production for E' allows for the possibility of ending the expression without further additions or multiplications.

This modified grammar is not left-recursive and can be used for parsing arithmetic expressions, where the E non-terminal represents expressions, T represents terms, and a and b represent basic elements in the language.

4. Construct the LALR parsing table for the given grammar

S →BB
B→ aB / b

Grammar:

S → BB B → aB B → b

Here are the LR(0) items for this grammar:

I0:
    S' → S-
    S → -BB
    B → -aB
    B → -b

I1:
    S' → S-
    S → B-B

I2:
    B → a-B
    B → -aB
    B → -b

I3:
    S → B-B

I4:
    B → b-

I5:
    B → -aB
    B → -b

Next, we merge the LR(0) items to create LALR(1) items. For simplicity, I0, I2, I4, and I5 merge to form one state, and I1 and I3 merge to form another state. We then compute the LALR(1) action and goto transitions for each state.

Here's the LALR parsing table:

State Action Goto
0 Shift to State 2
1 Shift to State 3
2 Shift to State 4
3 Accept
4 Reduce B → b
5 Reduce B → aB
  • State 0 corresponds to LR(0) items I0, I2, I4, and I5.
  • State 1 corresponds to LR(0) items I1 and I3.

5. Construct predictive parse table for the following grammar

E → E + T/T
T →T *F/F
F →F /a/b

Grammar:

E → E + T E → T T → T * F T → F F → F / a F → b

Let's construct the predictive parsing table. The table will have rows for non-terminals and columns for terminals and the end-of-input marker, '$'. Each table entry contains a production rule from the grammar.

Here's the predictive parsing table for the given grammar:

+ * / a b $
E E → E + T E → T
T T → T * F T → F T → F
F F → F / a F → b

The entries in the table are filled based on the production rules for each non-terminal when a specific terminal or '$' (end-of-input marker) is encountered. For example, if we are in state 'E' and encounter the '+' terminal, we use the production rule 'E → E + T'.

6. Differentiate Parse tree and Syntax tree with an example

Parse Tree:

A parse tree, also known as a derivation tree, is a tree-like graphical representation that shows the syntactic structure of a given input string based on a specific context-free grammar. It is used primarily during the parsing phase of a compiler to illustrate the application of grammar rules and the order in which terminals and non-terminals are generated.

A parse tree is often quite detailed and reflects the exact parsing process, including all the derivational steps. It's commonly used in the early stages of the compilation process.

Syntax Tree (Abstract Syntax Tree, or AST):

A syntax tree, also known as an abstract syntax tree (AST), is a more abstract and simplified representation of the syntactic structure of a program or input. It is commonly used in later phases of the compilation process, especially during semantic analysis and code generation. The syntax tree abstracts away many of the low-level details present in a parse tree and focuses on the essential structures and semantics of the program.

Here's an example to illustrate the difference between a parse tree and a syntax tree:

Input Expression:

a + b * c

Parse Tree: A parse tree for the input expression "a + b * c" might look like this:

+ /\ a * /\ b c

In this parse tree, every terminal and non-terminal is represented, and it reflects the exact sequence of derivational steps based on the grammar rules. It shows how the expression was parsed.

Syntax Tree (Abstract Syntax Tree - AST): An abstract syntax tree for the same input expression "a + b * c" might look like this:

+ /\ a * /\ b c

In this syntax tree (AST), the low-level details of the parsing process, such as non-terminals and unnecessary intermediate steps, are abstracted away. It focuses on the essential structure of the expression, emphasizing the operator precedence and the relationships between operands.

7. Test whether the grammar is LL(1) or not, and construct parsing table for it

S→1AB / ε
A→1AC/0C
B→0S
C→1

To test whether a grammar is LL(1), we need to check for two main properties:

  1. No Left Recursion: The grammar should not have left recursion. Left recursion is a situation where a non-terminal derives itself directly or indirectly through a series of productions.

  2. No Ambiguity: The grammar should not be ambiguous. Ambiguity occurs when the same input can be parsed in multiple ways, leading to uncertainty in parsing.

Let's analyze the given grammar:

Grammar:

S → 1AB | ε A → 1AC | 0C B → 0S C → 1

  1. No Left Recursion: The given grammar does not have any left recursion, so it satisfies the first property.

  2. No Ambiguity: To check for ambiguity, we need to examine the FIRST and FOLLOW sets for each non-terminal and the grammar productions. In an LL(1) grammar, there should be no multiple entries in the parsing table for the same input symbol.

Let's calculate the FIRST and FOLLOW sets:

FIRST sets:

  • FIRST(S) = {1, ε}
  • FIRST(A) = {0, 1}
  • FIRST(B) = {0}
  • FIRST(C) = {1}

FOLLOW sets:

  • FOLLOW(S) = {$, 0}
  • FOLLOW(A) = {0}
  • FOLLOW(B) = {1, $}
  • FOLLOW(C) = {0, $}

Now, let's construct the LL(1) parsing table:

0 1 $
S ε 1AB ε
A 0C 1AC
B 0S
C

The LL(1) parsing table is constructed by considering each non-terminal and its FIRST set, along with the terminals and the symbols that can be derived from them. There are no multiple entries for the same input symbol, which means that this grammar is LL(1).

8. Eliminate left recursion from the following grammar S→AB , A→BS | b , B→SA | a

To eliminate left recursion from the given grammar, you can follow these steps:

  1. Identify the non-terminals with left recursion.
  2. Create new non-terminals for each left-recursive non-terminal.
  3. Modify the productions to remove left recursion.

Let's apply these steps to the given grammar:

Original Grammar:

S → AB A → BS | b B → SA | a

  1. We can see that both non-terminals A and B have left recursion.

  2. Create new non-terminals A' and B' for A and B, respectively:

Intermediate Grammar:

S → AB A → A'S | b B → B'S | a A' → ε B' → ε

  1. Modify the productions to remove left recursion. To do this, you create two sets of productions for each non-terminal: one for the non-terminals with the left recursion and one for the new non-terminals. The ε productions are used to ensure that the original non-terminals can derive empty strings when needed:

Modified Grammar:

S → AB A → bA' A' → SA' | ε B → aB' B' → SB' | ε

The modified grammar is now free of left recursion. It maintains the same language as the original grammar but is more suitable for parsing without left-recursive issues.

9. Consider the following grammar E→E+E | E*E|(E)|id . Construct the SLR parsing table and suggest your final parsing table

To construct the SLR parsing table for the given grammar, we need to perform the following steps:

  1. Compute the FIRST and FOLLOW sets for each non-terminal in the grammar.

  2. Build the Canonical LR(0) Collection of LR(0) items for the grammar.

  3. Create the SLR(1) parsing table based on the Canonical LR(0) Collection.

Given Grammar:

mathematica

E → E + E | E * E | (E) | id

Step 1: Compute FIRST and FOLLOW sets:

I won't go through the details of computing FIRST and FOLLOW sets, but I'll provide the sets for the given grammar:

FIRST sets:

  • FIRST(E) = { (, id }
  • FIRST(+) = { + }
  • FIRST() = { }
  • FIRST(() = { ( }
  • FIRST()) = { ) }

FOLLOW sets:

  • FOLLOW(E) = { ), $, +, * }
  • FOLLOW(+) = { (, id }
  • FOLLOW(*) = { (, id }
  • FOLLOW(() = { (, id }
  • FOLLOW()) = { ), $, +, * }

Step 2: Build the Canonical LR(0) Collection:

I won't go through the details of constructing the Canonical LR(0) Collection, but it consists of LR(0) items representing possible states of a parser when parsing the grammar.

Step 3: Create the SLR(1) parsing table:

Using the Canonical LR(0) Collection, you can create the SLR(1) parsing table. I'll provide the parsing table without going through the process of construction:

SLR(1) Parsing Table:

State id + * ( ) $ E
0 s5 s4 1
1 s6 s7 acc
2 r4 r4 r4 r4 r4 r4
3 r5 r5 r5 r5 r5 r5
4 s5 s4 8
5 r7 r7 r7 r7 r7 r7
6 s5 s4 9
7 s5 s4 10
8 s6 s7 s11
9 r1 r1 r1 r1 r1 r1
10 r2 r2 r2 r2 r2 r2
11 r3 r3 r3 r3 r3 r3

In the parsing table:

  • sX denotes a shift operation to state X.
  • rX denotes a reduce operation using production rule X.
  • acc denotes the acceptance state.

10. Differentiate between LR and LL parser

Aspect LR Parser LL Parser
Type of Parsing Bottom-up parsing Top-down parsing
Parsing Strategy Uses a shift-reduce approach Uses a recursive descent approach
Grammar Types Suitable for a wide range of context-free grammars, including ambiguous grammars Generally works for a smaller class of context-free grammars, typically unambiguous
Table Building Constructs the LR(0), SLR(1), LALR(1), or LR(1) parsing tables Constructs the parsing table by examining FIRST and FOLLOW sets, which are used for LL(1) parsing
Left Recursion LR parsers can handle grammars with left recursion, but they may require left-factoring LL parsers often struggle with left-recursive grammars and typically require elimination
Precedence Can handle operator precedence and associativity through table entries and precedence declarations May require additional grammar rules or manual adjustments to handle operator precedence and associativity
Error Handling Typically better at providing detailed error messages and error recovery May provide less detailed error messages and can have limited error recovery
Performance Generally more efficient for parsing complex languages Can be less efficient for languages with complex or ambiguous syntax
Example Tools Bison, Yacc, ANTLR, LALR, etc. ANTLR, JavaCC, ANTLR, etc.

11. Differentiate between stack allocation and heap allocation

Aspect Stack Allocation Heap Allocation
Memory Management Area Utilizes the stack memory region in the program's memory space. Utilizes the heap memory region, which is a separate area from the stack.
Allocation/Deallocation Automatically allocated and deallocated, typically using a Last-In-First-Out (LIFO) mechanism. Requires explicit allocation and deallocation by the programmer, often manually.
Data Size Typically used for smaller, fixed-size data structures like local variables and function call frames. Suitable for dynamic data structures and objects of varying sizes, including large data.
Speed Faster allocation and deallocation since it involves a simple adjustment of the stack pointer. Slower allocation and deallocation due to the need for memory management and tracking.
Lifetime Limited to the scope of the block or function where variables are declared. Variables can have a longer or dynamic lifetime, extending beyond the current scope.
Memory Fragmentation Rarely suffers from memory fragmentation. Prone to memory fragmentation, especially when memory is allocated and deallocated frequently.
Examples Used for local variables in functions, function call stacks, and automatic memory management. Used for dynamic data structures like arrays, objects, and memory allocated using functions like malloc or new.

12. Consider the following three address code segments

PROD := 0
I:= 1
T1:=4*I
T2:=addr(A)‐4
T3:=T2[T1]
T4:=addr(B)‐4
T5:=T4[T1]
T6:=T3*T5
PROD:=PROD +T6
I:=I+1
If i<=20 goto (3)
a. Find the basic blocks and flow graph of above 
sequence.
b. Optimize the code sequence by applying function 
preserving transformation and loop optimization 
technique.

a. Finding Basic Blocks and Flow Graph:

To find the basic blocks, we need to identify sequences of code that have a single entry point and a single exit point. The flow graph connects these basic blocks based on control flow instructions (e.g., conditional branches or jumps). Here are the basic blocks and the flow graph for the provided code:

Basic Blocks:

  1. PROD := 0
  2. I := 1
  3. T1 := 4 * I
  4. T2 := addr(A) - 4
  5. T3 := T2[T1]
  6. T4 := addr(B) - 4
  7. T5 := T4[T1]
  8. T6 := T3 * T5
  9. PROD := PROD + T6
  10. I := I + 1
  11. If I <= 20 goto (3)

Flow Graph:

  • Basic Blocks 1, 2, and 3 are executed sequentially without any branches. So, they form a single basic block.
  • Basic Block 4 is a separate block due to the assignment operation.
  • Basic Blocks 5 and 6 are also separate blocks.
  • Basic Blocks 7, 8, and 9 are sequential and contain arithmetic operations, so they form a single block.
  • Basic Blocks 10 and 11 are separate due to the assignment operation.
  • The conditional branch If I <= 20 goto (3) forms a loop by connecting Basic Blocks 11 and 3.

Flow Graph:

    [1,2,3]
     / |\
    /  | \
   /   |  \
  4    |   11
   \   |  /
    \  | /
     \ |/
     [5,6]
       |
    [7,8,9]
       |
      [10]

b. Optimizing the Code:

To optimize the provided code, we can apply function-preserving transformations and loop optimizations. In this case, we can see that there's a loop from Basic Block 11 to Basic Block 3. We can apply loop optimization techniques like loop unrolling to improve performance. Here's an optimized version of the code:

PROD := 0
I := 1

// Unrolled loop for 20 iterations
T1 := 4 * I
T2 := addr(A) - 4
T3 := T2[T1]
T4 := addr(B) - 4
T5 := T4[T1]
T6 := T3 * T5
PROD := PROD + T6
I := I + 1

T1 := 4 * I
T2 := addr(A) - 4
T3 := T2[T1]
T4 := addr(B) - 4
T5 := T4[T1]
T6 := T3 * T5
PROD := PROD + T6
I := I + 1

// Repeat the unrolled loop for a total of 20 iterations
// ...

In the optimized version, we've unrolled the loop to reduce the number of conditional branches and improve performance. This is just one example of optimization, and other techniques can also be applied depending on the specific requirements and constraints of your target platform.

13. What are the advantages of DAG? Explain the peephole optimization

Directed Acyclic Graph (DAG):

A Directed Acyclic Graph (DAG) is a data structure used in compilers and various optimization techniques. It represents a program's control flow or data flow in a way that helps optimize code generation and analysis. DAGs have several advantages, including:

  1. Optimization: One of the primary advantages of a DAG is that it can reveal opportunities for optimization by identifying common subexpressions and redundant calculations in a program. These optimizations include common subexpression elimination, constant folding, and strength reduction.

  2. Reduced Redundancy: DAGs help reduce redundancy in the intermediate code by identifying and eliminating redundant subexpressions. This, in turn, can lead to more efficient code generation.

  3. Improved Code Quality: The optimization techniques applied using DAGs can lead to improved code quality, with reduced execution time and memory usage.

Peephole Optimization:

Peephole optimization is a local optimization technique that focuses on a small, contiguous sequence of instructions in the compiled code (the "peephole") and looks for opportunities to make the code more efficient. It operates on the machine code or assembly code level, examining a short window of instructions. The primary goal of peephole optimization is to eliminate or simplify redundant or inefficient code patterns.

Key characteristics of peephole optimization include:

  • Local Scope: Peephole optimization operates within a limited scope, usually a small number of contiguous instructions. It does not consider the entire program's context.

  • Pattern Matching: It identifies specific code patterns that can be replaced with more efficient or simplified alternatives. For example, replacing a sequence of instructions with a single instruction that achieves the same result.

  • No Structural Changes: Peephole optimization does not alter the control flow or structure of the program; it focuses solely on improving the code within the given window.

Common peephole optimization techniques include:

  1. Constant Folding: Replacing constant expressions with their computed values. For example, replacing 3 * 4 with 12.

  2. Dead Code Elimination: Removing code that has no effect on the program's behavior or result. This includes eliminating unreachable code or code that computes values that are never used.

  3. Strength Reduction: Replacing expensive operations with cheaper ones. For example, replacing division by a constant with multiplication by its reciprocal.

  4. Common Subexpression Elimination: Identifying and removing redundant computations by reusing the result of previous calculations.

14. Write short note on

  • Loop optimization
  • Global data analysis
  • Direct acyclic graph
  • YACC parser generator

i. Loop Optimization: Loop optimization is a critical phase in compiler optimization that focuses on improving the performance of loops in a program. Loops are a fundamental control structure in many programs, and optimizing them can lead to substantial performance improvements. Common loop optimizations include loop unrolling (generating multiple copies of loop bodies to reduce loop control overhead), loop fusion (combining multiple loops into one to reduce memory access), and loop parallelization (running loop iterations in parallel to take advantage of multiple processors).

ii. Global Data Analysis: Global data analysis, often referred to as data flow analysis, is a compiler optimization technique that analyzes how data values propagate through a program. It helps identify the relationships between variables and how they are used, which is crucial for optimizing code. Global data analysis can be used to detect opportunities for common subexpression elimination, dead code elimination, and other optimizations that improve the efficiency of a program.

iii. Directed Acyclic Graph (DAG): A Directed Acyclic Graph (DAG) is a data structure used in compilers and optimization techniques. It represents the control flow or data flow of a program, revealing opportunities for optimization by identifying common subexpressions and redundant calculations. DAGs are particularly useful for optimizing code generation and analysis, including common subexpression elimination, constant folding, and strength reduction.

iv. YACC Parser Generator: YACC (Yet Another Compiler Compiler) is a tool used for generating parsers and syntax analyzers for programming languages. It takes a formal grammar as input and generates parser code in a programming language (typically C or C++). YACC helps automate the process of building parsers, making it easier to develop compilers and interpreters for new programming languages. YACC-based parsers use context-free grammars to parse and analyze the syntax of a program and generate abstract syntax trees for further processing. YACC is a valuable tool in compiler construction and language design.

15. Explain the following

  • Copy Propagation
  • Dead-Code Elimination
  • Code Motion
  • Reduction in Strength

(i) Copy Propagation: Copy propagation is a compiler optimization technique that replaces the use of a variable with the value it was assigned, whenever possible. This optimization is based on the concept that if a variable's value is assigned to another variable (a copy), then the two variables are equivalent in terms of their values within a specific scope. Copy propagation reduces the number of variable accesses, leading to more efficient code. For example, if x is assigned the value of y, and later, x is used, copy propagation would replace x with y. This optimization simplifies the code and may lead to further opportunities for optimization.

(ii) Dead-Code Elimination: Dead-code elimination is an optimization technique that identifies and removes code that has no effect on a program's behavior or output. Dead code typically includes statements or variables that are never used or have no impact on the program's final result. Eliminating dead code can lead to more efficient and compact programs, as it reduces the computational and memory overhead associated with unnecessary instructions. Dead-code elimination is essential in optimizing code and reducing the program's footprint.

(iii) Code Motion: Code motion, often referred to as loop-invariant code motion (LICM), is a compiler optimization that identifies expressions or instructions within loops that do not change their values across loop iterations. These loop-invariant computations are moved outside the loop to reduce redundant calculations. Code motion can improve the efficiency of loops by executing expensive operations only once. It's an essential optimization, particularly for improving the performance of loops and reducing execution time.

(iv) Reduction in Strength: Reduction in strength is a compiler optimization technique that replaces expensive operations with cheaper alternatives to achieve the same result. This optimization aims to reduce the computational cost of instructions. For example, replacing division operations with multiplication by the reciprocal, or converting expensive floating-point operations into integer operations. Reduction in strength is essential for improving the efficiency of code, especially in numeric or computational applications, where performance is critical. This optimization helps to optimize the trade-off between accuracy and speed in numeric computations.

16. Explain in the DAG representation of the basic block with example

A Directed Acyclic Graph (DAG) representation of a basic block is a data structure that depicts the control and data flow within a sequence of instructions, often within a single basic block of a program or function. The DAG is particularly useful for optimizing code by identifying common subexpressions and dependencies between instructions.

Let's look at an example to illustrate the DAG representation of a basic block:

Consider the following basic block of code:

t1 = a + b t2 = a * c t3 = t1 - t2 d = t3 * 2

Now, let's create a DAG representation of this basic block:

In this representation:

  • Each variable assignment (e.g., t1 = a + b) is considered a node in the DAG.
  • The nodes are connected by directed edges to represent the flow of data from one instruction to another.
  • Common subexpressions are identified and shared as much as possible to reduce redundancy.

The DAG for the given basic block might look like this:

     +
        /\
       /\
      t1   *
          /\
         /\
        a     c\
          -
         /\
        /\
       t3    t2\
                *
               /\
              /\
             2     t3\
                    *
                   /\
                  /\
                 d     2

In this DAG representation:

  • t1 and t2 are common subexpressions shared between t3 = t1 - t2 and d = t3 * 2.
  • The nodes for variables a, b, c, and d are leaves, as they are the inputs to the expressions.
  • The nodes for the operators (+, *, and -) represent the computation being performed.

This DAG allows the compiler to identify and optimize common subexpressions, potentially eliminating redundant calculations. For example, the value of t3 is computed only once, and it's reused in the calculation of d. By recognizing such opportunities for optimization, the compiler can produce more efficient code.

17. Differentiate between compiler, interpreter and assembler

Aspect Compiler Interpreter Assembler
Input Language Translates high-level source code into machine code or an intermediate language. Executes high-level source code directly, line by line, without generating an intermediate language. Translates assembly language into machine code.
Output Generates machine code or an executable file. Does not generate machine code; it directly executes the source code. Produces object code or machine code that can be linked to create an executable program.
Execution Speed Usually results in faster execution because the source code is fully translated before execution. Generally slower in execution as it interprets code line by line, without a complete translation. Execution speed varies but is typically faster than high-level source code due to closer proximity to machine code.
Error Handling Error detection and reporting may occur during the compilation process, with a list of errors generated. Errors are often detected and reported as the code is executed, stopping execution upon encountering an error. Errors are reported during assembly, and debugging can be more challenging than with high-level languages.
Portability Output code is usually platform-dependent, requiring recompilation for different platforms. Source code can be platform-independent, but the interpreter itself must be platform-specific. Output code is platform-dependent, and minor changes may be needed for different architectures.
Examples C/C++, Java, C#, etc. Python, Ruby, JavaScript, etc. x86 assembly, MIPS assembly, etc.

18. Discuss how induction variables can be detected and eliminated from the given intermediate code

B2: i = i+1
t1: = 4*j
t2: = a[t1]
if t2 < 10 goto B2

Induction variables are variables whose values change by a constant amount in each iteration of a loop. They are often used as loop control variables, and optimizing them can lead to performance improvements. In your given intermediate code, you have a loop that increments an induction variable i and uses it to index an array. We will discuss how to detect and potentially eliminate this induction variable:

Intermediate Code:

B2: i = i + 1
t1: = 4 * j
t2: = a[t1]
if t2 < 10 goto B2

To detect and potentially eliminate the induction variable i, you need to analyze the code and the control flow within the loop. In this case, i is incremented by 1 in each iteration, and its value is used to index the array a. Here are the steps:

  1. Detecting the Induction Variable:

    • Identify variables that are incremented by a constant value in each iteration. In this case, i is incremented by 1 in every loop iteration.

    • Analyzing Loop Properties:

    • Determine whether the loop is well-structured and can be safely optimized. In this case, it appears to be a simple incrementing loop with a termination condition (t2 < 10).

    • Optimization Opportunities:

    • Since i is an induction variable, it can potentially be eliminated by rewriting the loop using a different approach. In this case, you might rewrite the loop without explicitly using i for indexing. For example, you can use a loop counter to access the array.

Here's an example of how you can rewrite the loop to eliminate the use of i:

loop_counter = 0
t1 = 0
while loop_counter < n:
    t2 = a[t1]  # Access the array using t1
    if t2 < 10:
        loop_counter = loop_counter + 1
        t1 = t1 + 4  # Increment t1 instead of i

By eliminating the explicit use of i, you reduce the number of instructions and potentially improve code efficiency. However, the exact optimization may depend on the context and requirements of the program. Additionally, the compiler may perform such optimizations automatically as part of loop optimization techniques.

19. Construct the flow graph for the following code segment

fact(n)
{
int f=1;
for(i=2; i≤n; i++)
f=f*i;
return f;
}

To construct the flow graph for the given code segment, we'll create a representation of the control flow within the fact function. The code includes a simple loop that calculates the factorial of a number n. Here's the flow graph:

Start
|
v Assignment: f = 1
|
v Assignment: i = 2
|
v Condition: i <= n?
|
| (no)
v Return: f
|
v Condition: i <= n?
|
| (yes)
v Assignment: f = f * i
|
v Assignment: i = i + 1
|
v Back to Condition

Explanation of the flow graph:

  • The flow graph starts with the "Start" node.
  • It then proceeds to the assignment of f = 1.
  • Next, it assigns i = 2.
  • The first condition checks whether i is less than or equal to n. If it's not, the control flow exits the loop, proceeds to the "Return: f" node, and eventually exits the function.
  • If the condition is true, indicating that the loop should continue, the control flow proceeds to the assignment f = f * i.
  • After that, it assigns i = i + 1 to increment the loop counter.
  • Finally, the control flow goes back to the condition, where the loop continues until i is no longer less than or equal to n.