All questions of Compiler Design for Computer Science Engineering (CSE) Exam
'Divide by Zero ’ is a- a)lexical error
- b)syntax error
- c)semantic error
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
'Divide by Zero ’ is a
a)
lexical error
b)
syntax error
c)
semantic error
d)
none of these
|
Bijoy Sharma answered |
It is a semantic error (run-time error).
Which of the following actions an operator- precedence parser may take to recover from an error?- a)Insert symbols onto the stack
- b)Delete symbols from the stack
- c)Insert or delete symbols from the input
- d)All of the above
Correct answer is option 'D'. Can you explain this answer?
Which of the following actions an operator- precedence parser may take to recover from an error?
a)
Insert symbols onto the stack
b)
Delete symbols from the stack
c)
Insert or delete symbols from the input
d)
All of the above
|
Sanya Agarwal answered |
To recover from an error actions taken by operator precedence parser:
1. Insert symbols onto stack.
2. Delete symbols from the stack.
3. Insert or delete symbols from the input.
1. Insert symbols onto stack.
2. Delete symbols from the stack.
3. Insert or delete symbols from the input.
Which languages necessarily need heap allocation in the runtime environment?- a)Those that support recursion .
- b)Those that use dynamic scoping
- c)Those that allow dynamic data structure
- d)Those that use global variables
Correct answer is option 'C'. Can you explain this answer?
Which languages necessarily need heap allocation in the runtime environment?
a)
Those that support recursion .
b)
Those that use dynamic scoping
c)
Those that allow dynamic data structure
d)
Those that use global variables
|
|
Sanya Agarwal answered |
Runtime environment means we deal with dynamic memory allocation and Heap is a dynamic data structure.
So it is clear that those languages that allow dynamic data structure necessarily need heap allocation in the runtime environment.
So it is clear that those languages that allow dynamic data structure necessarily need heap allocation in the runtime environment.
The term “environment” in programming language semantics is said as- a)Function that maps a name to value held there.
- b)Function that maps a name to a storage location.
- c)The function that maps a storage location to the value held there.
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
The term “environment” in programming language semantics is said as
a)
Function that maps a name to value held there.
b)
Function that maps a name to a storage location.
c)
The function that maps a storage location to the value held there.
d)
None of the above
|
Ananya Kumari answered |
The term environment in programming language semantic is said as
function that maps a name to a stroage location.
function that maps a name to a stroage location.
Which statement is true?- a)LALR parser is most powerful and costly as compare to other parsers.
- b)All CFG’s are LR and not all grammars are uniquely defined.
- c)Every SLR grammar is unambiguous but not every unambiguous grammar is SLR.
- d)LR(K) is the most general back tracking shift reduce parsing method.
Correct answer is option 'C'. Can you explain this answer?
Which statement is true?
a)
LALR parser is most powerful and costly as compare to other parsers.
b)
All CFG’s are LR and not all grammars are uniquely defined.
c)
Every SLR grammar is unambiguous but not every unambiguous grammar is SLR.
d)
LR(K) is the most general back tracking shift reduce parsing method.
|
Rohan Shah answered |
- LALR is less powerful than CLR but more costly than any parser.
- Every SLR grammar is unambigous but every unambigous grammar need not be SLR i.e. may be LALR but not SLR.
- LR(K) grammar never using backtracking.
Pick the functions which are completely performed in pass 1- a)Updating the location counter
- b)Processing of EQU pseudo op
- c)Processing of DS pseudo op
- d)Ail of the above
Correct answer is option 'D'. Can you explain this answer?
Pick the functions which are completely performed in pass 1
a)
Updating the location counter
b)
Processing of EQU pseudo op
c)
Processing of DS pseudo op
d)
Ail of the above
|
Jatin Joshi answered |
Operation performed during pass-1 in 2-pass compiler:
1. Processing the DB pseudo-op.
2. Updating the location counter.
3. Processing of EQU pseudo-op.
4. Processing of DS pseudo-op.
1. Processing the DB pseudo-op.
2. Updating the location counter.
3. Processing of EQU pseudo-op.
4. Processing of DS pseudo-op.
In an incompletely specified automata- a)No edge should be labeled epsilon.
- b)From any given state, there can’t be any token leading to two different states.
- c)Some states have no transition on some tokens.
- d)Start state may not be there.
Correct answer is option 'C'. Can you explain this answer?
In an incompletely specified automata
a)
No edge should be labeled epsilon.
b)
From any given state, there can’t be any token leading to two different states.
c)
Some states have no transition on some tokens.
d)
Start state may not be there.
|
Avantika Yadav answered |
In incomplete automata some states have no transition on some tokens.
MIMD stands for __________- a)Multiple instruction multiple data
- b)Multiple instruction memory data
- c)Memory instruction multiple data
- d)Multiple information memory data
Correct answer is option 'A'. Can you explain this answer?
MIMD stands for __________
a)
Multiple instruction multiple data
b)
Multiple instruction memory data
c)
Memory instruction multiple data
d)
Multiple information memory data
|
Abhiram Goyal answered |
MIMD stands for Multiple Instruction Multiple Data.
Multiple Instruction Multiple Data (MIMD) is a type of parallel processing architecture in which multiple processors or cores execute different instructions on different data simultaneously. Each processor or core has its own instruction stream and data stream, allowing for independent execution of instructions.
Explanation:
MIMD is a parallel computing architecture that utilizes multiple processors or cores to execute different instructions on different data simultaneously. It is designed to achieve high-performance computing by dividing the workload among multiple processors and allowing them to work in parallel.
In MIMD architecture, each processor or core operates independently and has its own instruction stream and data stream. This means that each processor can execute different instructions on different data at the same time. This allows for simultaneous execution of multiple instructions, leading to faster processing and increased efficiency.
Advantages of MIMD architecture:
- Increased performance: MIMD architecture allows for parallel execution of multiple instructions on different data, which leads to faster processing and increased performance.
- Scalability: MIMD architecture can be scaled up by adding more processors or cores to the system, allowing for increased processing power as the workload demands.
- Flexibility: Each processor or core in a MIMD system can execute different instructions, making it suitable for a wide range of applications that require different types of computations.
- Fault tolerance: MIMD architecture provides redundancy by having multiple processors or cores. If one processor fails, the other processors can continue to operate, ensuring the system's reliability.
Examples of MIMD architecture:
- Multi-core processors: Modern CPUs often have multiple cores that can execute different instructions on different data simultaneously, making them MIMD architectures.
- Distributed computing systems: Systems that utilize multiple computers or servers to perform parallel processing tasks are also examples of MIMD architecture. Each computer or server can execute different instructions on different data, allowing for parallel processing.
In conclusion, MIMD stands for Multiple Instruction Multiple Data, which is a parallel computing architecture that utilizes multiple processors or cores to execute different instructions on different data simultaneously. This architecture provides increased performance, scalability, and flexibility, making it suitable for a wide range of applications.
Multiple Instruction Multiple Data (MIMD) is a type of parallel processing architecture in which multiple processors or cores execute different instructions on different data simultaneously. Each processor or core has its own instruction stream and data stream, allowing for independent execution of instructions.
Explanation:
MIMD is a parallel computing architecture that utilizes multiple processors or cores to execute different instructions on different data simultaneously. It is designed to achieve high-performance computing by dividing the workload among multiple processors and allowing them to work in parallel.
In MIMD architecture, each processor or core operates independently and has its own instruction stream and data stream. This means that each processor can execute different instructions on different data at the same time. This allows for simultaneous execution of multiple instructions, leading to faster processing and increased efficiency.
Advantages of MIMD architecture:
- Increased performance: MIMD architecture allows for parallel execution of multiple instructions on different data, which leads to faster processing and increased performance.
- Scalability: MIMD architecture can be scaled up by adding more processors or cores to the system, allowing for increased processing power as the workload demands.
- Flexibility: Each processor or core in a MIMD system can execute different instructions, making it suitable for a wide range of applications that require different types of computations.
- Fault tolerance: MIMD architecture provides redundancy by having multiple processors or cores. If one processor fails, the other processors can continue to operate, ensuring the system's reliability.
Examples of MIMD architecture:
- Multi-core processors: Modern CPUs often have multiple cores that can execute different instructions on different data simultaneously, making them MIMD architectures.
- Distributed computing systems: Systems that utilize multiple computers or servers to perform parallel processing tasks are also examples of MIMD architecture. Each computer or server can execute different instructions on different data, allowing for parallel processing.
In conclusion, MIMD stands for Multiple Instruction Multiple Data, which is a parallel computing architecture that utilizes multiple processors or cores to execute different instructions on different data simultaneously. This architecture provides increased performance, scalability, and flexibility, making it suitable for a wide range of applications.
Resolution is externally defined symbols is performed by- a)Linker
- b)Loader
- c)Compiler
- d)Assembler
Correct answer is option 'A'. Can you explain this answer?
Resolution is externally defined symbols is performed by
a)
Linker
b)
Loader
c)
Compiler
d)
Assembler
|
Athul Pillai answered |
Resolution is generally done during linking. Hence, linker does this
Consider line number 3 of the following C-program.
int main () { /* Line 1 */
int i, n; /* Line 2 */
fro (i = 0, i < n, /+ + ); /* Line 3 */
}Identify the compiler’s response about this line while creating the object - module- a)No compilation error
- b)Only a lexical error
- c)Only syntactic errors
- d)Both lexical and syntactic errors
Correct answer is option 'C'. Can you explain this answer?
Consider line number 3 of the following C-program.
int main () { /* Line 1 */
int i, n; /* Line 2 */
fro (i = 0, i < n, /+ + ); /* Line 3 */
}
int main () { /* Line 1 */
int i, n; /* Line 2 */
fro (i = 0, i < n, /+ + ); /* Line 3 */
}
Identify the compiler’s response about this line while creating the object - module
a)
No compilation error
b)
Only a lexical error
c)
Only syntactic errors
d)
Both lexical and syntactic errors
|
Navya Menon answered |
Error is underlined function fro()
Which is a syntactical error rather than lexical.
Which is a syntactical error rather than lexical.
Three address code’ technique for intermediate code generation shows that each statement usually contains three addresses. Three addresses are as follows:- a)Two addresses for operands, one for operator
- b)One for all operator, one for all operands and one for result
- c)One for result, two for operands
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Three address code’ technique for intermediate code generation shows that each statement usually contains three addresses. Three addresses are as follows:
a)
Two addresses for operands, one for operator
b)
One for all operator, one for all operands and one for result
c)
One for result, two for operands
d)
None of the above
|
Garima Bose answered |
Three address code formate is two addresses for operands and one for operator.
Choose the false statements:
1. Control stack keeps track of live procedure activations.
2. Activation records can be managed with the help of stack.
3. Dangling reference is a reference to a storage that has been deallocated- a)1 and 2
- b)2 and 3
- c)1 and 3
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Choose the false statements:
1. Control stack keeps track of live procedure activations.
2. Activation records can be managed with the help of stack.
3. Dangling reference is a reference to a storage that has been deallocated
1. Control stack keeps track of live procedure activations.
2. Activation records can be managed with the help of stack.
3. Dangling reference is a reference to a storage that has been deallocated
a)
1 and 2
b)
2 and 3
c)
1 and 3
d)
None of these
|
|
Vaishnavi Dey answered |
- Control stack keeps tracks of live procedure.
- Activation records can be managed with.
- Dangling reference is a reference to a storage that has been de-allocated.
All statement are true.
Which of the following are language processors?- a)Assembler
- b)compilers
- c)Interpreters
- d)All of these
Correct answer is option 'A'. Can you explain this answer?
Which of the following are language processors?
a)
Assembler
b)
compilers
c)
Interpreters
d)
All of these
|
Mira Rane answered |
Assembler is a language processor to process assembly language
in a context-free grammar- a)∈ can’t be the right hand side of any production.
- b)Terminal symbols can’t be present in the left hand side of any production.
- c)The number of grammar symbols in the left hand side is not greater than that of in the right hand side.
- d)All of the above
Correct answer is option 'B'. Can you explain this answer?
in a context-free grammar
a)
∈ can’t be the right hand side of any production.
b)
Terminal symbols can’t be present in the left hand side of any production.
c)
The number of grammar symbols in the left hand side is not greater than that of in the right hand side.
d)
All of the above
|
|
Megha Yadav answered |
The context free grammar is represented as
S → X
S → X
where S∈ V(Single symbol non-terminal)
D → (V ∪ T ) * (Any number of terminal and nonterminals)
A basic block can be analyzed by- a)DAG
- b)A flow graph
- c)A graph with cycles
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
A basic block can be analyzed by
a)
DAG
b)
A flow graph
c)
A graph with cycles
d)
None of the above
|
Amrutha Singh answered |
Properties of a basic block are as follows:
- The flow of control can only enter the basic block through the first instruction in the block.
- - No numps in the middle of the block.
- - Control leaves the block without halting/branching.
- - The basic blocks become the nodes of a flow graph, whose edges indicate which blocks can follow which other blocks.
Hence Option (A) is correct
For video on Intermediate code generation click on the link given below:
For a context-free grammar, FOLLOW(A) is the set of terminals that can appear immediately to the right of non-terminal A in some “sentential” form. We defined to sets LFOLLOW(A) and RFOLLOW(A) by replacing the word “left most sentential” and “right most sentential” respectively in the definition of FOLLOW(A).Which of the following statements is/are tru- a)FOLLOW(A) and RFOLLOW(A) may be different.
- b)FOLLOW(A) and RFOLLOW(A) are always the same.
- c)All the three sets are identical.
- d)All the three sets are different.
Correct answer is option 'C'. Can you explain this answer?
For a context-free grammar, FOLLOW(A) is the set of terminals that can appear immediately to the right of non-terminal A in some “sentential” form. We defined to sets LFOLLOW(A) and RFOLLOW(A) by replacing the word “left most sentential” and “right most sentential” respectively in the definition of FOLLOW(A).
Which of the following statements is/are tru
a)
FOLLOW(A) and RFOLLOW(A) may be different.
b)
FOLLOW(A) and RFOLLOW(A) are always the same.
c)
All the three sets are identical.
d)
All the three sets are different.
|
Sarthak Desai answered |
Definition of Follow (A) as per question “The set of terminals that can appears immediately to the right of non-terminal A in same sentential form”. If the above definition is changed to replace “ sentential” with “ left-most sentential” for L FOLLOW (A) and “right-most sentential” for R FOLLOW (A)
We are still considering the terminal symbol to the right of A ’ .
Hence all three sets would be identical
Hence all three sets would be identical
The translator is best described as- a)Application software
- b)A system software
- c)A hardware component
- d)All of the above
Correct answer is option 'B'. Can you explain this answer?
The translator is best described as
a)
Application software
b)
A system software
c)
A hardware component
d)
All of the above
|
Samridhi Joshi answered |
Translator is a system software. Generally, it comes with the operating system.
Consider the following issues:
1. Simplify the phases.
2. Compiler efficiency is improved.
3. Compiler works faster.
4. Compiler portability is enhanced.Which is/are true in context of lexical analysis?- a)1,2 and 3
- b)1,3 and 4
- c)1,2 and 4
- d)All of these
Correct answer is option 'C'. Can you explain this answer?
Consider the following issues:
1. Simplify the phases.
2. Compiler efficiency is improved.
3. Compiler works faster.
4. Compiler portability is enhanced.
1. Simplify the phases.
2. Compiler efficiency is improved.
3. Compiler works faster.
4. Compiler portability is enhanced.
Which is/are true in context of lexical analysis?
a)
1,2 and 3
b)
1,3 and 4
c)
1,2 and 4
d)
All of these
|
Hiral Nair answered |
Lexical analysis consists of the phases that
(i) Simplify the subsequent phases by the help of tokens instead of arbitrary elements.
(ii) Since interrelated elements are converted to stream of tokens hence compiler efficiency is improved as no arbitrary storage or value is to be checked independently.
(iii) Computer portability is enhanced.
(i) Simplify the subsequent phases by the help of tokens instead of arbitrary elements.
(ii) Since interrelated elements are converted to stream of tokens hence compiler efficiency is improved as no arbitrary storage or value is to be checked independently.
(iii) Computer portability is enhanced.
A non relocatable program is one which- a)Cannot be made to execute in any area of storage other than the one designated for it at the time of its coding or translation.
- b)Consists of a program and relevant information for its relocation.
- c)Can itself perform the relocation of its address sensitive portions.
- d)All of the above
Correct answer is option 'A'. Can you explain this answer?
A non relocatable program is one which
a)
Cannot be made to execute in any area of storage other than the one designated for it at the time of its coding or translation.
b)
Consists of a program and relevant information for its relocation.
c)
Can itself perform the relocation of its address sensitive portions.
d)
All of the above
|
|
Gargi Menon answered |
The phases of compiler
1. Lexical
2. Syntax
3. Semantic
4. Intermediate code generation, are all machine dependent phases and are not relocatable.
The together make pass-1 of the compiler. This program hence can not be made to execute in any area of storage other than the one designated for it at the time of its coding or translation.
1. Lexical
2. Syntax
3. Semantic
4. Intermediate code generation, are all machine dependent phases and are not relocatable.
The together make pass-1 of the compiler. This program hence can not be made to execute in any area of storage other than the one designated for it at the time of its coding or translation.
In a bottom-up evaluation of a syntax directed definition, inherited attributes can- a)Always be evaluated
- b)Be evaluated only if the definition is L-attributed
- c)Be evaluated only if the definition has synthesized attributes
- d)Never be evaluated
Correct answer is option 'C'. Can you explain this answer?
In a bottom-up evaluation of a syntax directed definition, inherited attributes can
a)
Always be evaluated
b)
Be evaluated only if the definition is L-attributed
c)
Be evaluated only if the definition has synthesized attributes
d)
Never be evaluated
|
|
Rajat Sharma answered |
Every S(Synthesized) - attributed definition is L- attributed. For implementing inherited attributed during bottom-up parsing, extends to some, but not LR grammars. Consider the following example
In the example above the nonterminal L in L → E inherits the count of the number of 1 ’s generated by S. Since the production L → E is the first that a bottom- up parser would reduce by, the translator at the time can't know the number of 1 ’s in the input. So in a bottom-up evaluation of a syntax directed definition, inherits attributes can’t be evaluated if the definition is L-attributed in the given example. So we can say. that L-attributed definition is based on simple LR(1) grammar, but it can’t be implemented always but inherit attributes can be evaluated only if the definition has synthesized attributes.
In the example above the nonterminal L in L → E inherits the count of the number of 1 ’s generated by S. Since the production L → E is the first that a bottom- up parser would reduce by, the translator at the time can't know the number of 1 ’s in the input. So in a bottom-up evaluation of a syntax directed definition, inherits attributes can’t be evaluated if the definition is L-attributed in the given example. So we can say. that L-attributed definition is based on simple LR(1) grammar, but it can’t be implemented always but inherit attributes can be evaluated only if the definition has synthesized attributes.
The load instruction is mostly used to designate a transfer from memory to a processor register known as_______- a)Accumulator
- b)Instruction Register
- c)Program counter
- d)Memory address Register
Correct answer is option 'A'. Can you explain this answer?
The load instruction is mostly used to designate a transfer from memory to a processor register known as_______
a)
Accumulator
b)
Instruction Register
c)
Program counter
d)
Memory address Register
|
|
Sonal Nair answered |
Accumulator is the process register.
he average time required to reach a storage location in memory and obtain its contents is called__________- a)Latency time
- b)Access time
- c)Turnaround time
- d)Response time
Correct answer is option 'B'. Can you explain this answer?
he average time required to reach a storage location in memory and obtain its contents is called__________
a)
Latency time
b)
Access time
c)
Turnaround time
d)
Response time
|
Kiran Chakraborty answered |
Access time is the time from the start of one storage device access to the time when the next access can be started.
Which of the following derivations does a top- down parser use while parsing as input string? The input is assumed to be scanned in left to right order- a)Left most derivation
- b)Left most derivation traced out in reverse
- c)Right most derivation
- d)Right most derivation traced out in reverse
Correct answer is option 'A'. Can you explain this answer?
Which of the following derivations does a top- down parser use while parsing as input string? The input is assumed to be scanned in left to right order
a)
Left most derivation
b)
Left most derivation traced out in reverse
c)
Right most derivation
d)
Right most derivation traced out in reverse
|
Aaditya Ghosh answered |
Top down parser uses left to right, left most derivation.
In analyzing the compilation of PL/I program, the term “lexical analysis” is associated with- a)Recognition of basic syntactic constructs through reductions.
- b)Recognition of basic elements and creation of uniform symbols.
- c)Creation of more optimal matrix.
- d)Use of macro processor to produce more optimal assembly code.
Correct answer is option 'B'. Can you explain this answer?
In analyzing the compilation of PL/I program, the term “lexical analysis” is associated with
a)
Recognition of basic syntactic constructs through reductions.
b)
Recognition of basic elements and creation of uniform symbols.
c)
Creation of more optimal matrix.
d)
Use of macro processor to produce more optimal assembly code.
|
|
Kunal Sen answered |
The main task of lexical analyser is to read the input characters and producers as output sequence of tokens that the parser uses for syntax analysis.
Which of the following describes a handle (as applicable to LR-parsing) appropriately?- a)It is the position in a sentential form where the next shift or reduce operation will occur.
- b)It is a non-terminal whose production will be used for reduction in the next step.
- c)It is a production that may be used for reduction in a future step along with a position in the sentential form where the next shift or reduce operation will occur.
- d)It is the RHS of production p that will be used for reduction in the next step along with a position in the sentential.
Correct answer is option 'D'. Can you explain this answer?
Which of the following describes a handle (as applicable to LR-parsing) appropriately?
a)
It is the position in a sentential form where the next shift or reduce operation will occur.
b)
It is a non-terminal whose production will be used for reduction in the next step.
c)
It is a production that may be used for reduction in a future step along with a position in the sentential form where the next shift or reduce operation will occur.
d)
It is the RHS of production p that will be used for reduction in the next step along with a position in the sentential.
|
|
Aman Menon answered |
A handle may be described as the sentencial form given by the bottom up parser from left to right in reverse order.
If conversion from one type to another type is done automatically by the compiler then, it is called- a)Implicit conversion
- b)Coercions
- c)Both (a) and (b)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
If conversion from one type to another type is done automatically by the compiler then, it is called
a)
Implicit conversion
b)
Coercions
c)
Both (a) and (b)
d)
None of these
|
|
Nishanth Roy answered |
If compiler automatically converts one type of from to another type then it is called implicit conversion or coercions, else if not done automatically, then it is called explicit conversion.
A floating point number that has an O in the MSB of mantissa is said to have__________- a)Overflow
- b)Underflow
- c)Important number
- d)Undefined
Correct answer is option 'B'. Can you explain this answer?
A floating point number that has an O in the MSB of mantissa is said to have__________
a)
Overflow
b)
Underflow
c)
Important number
d)
Undefined
|
Raj Datta answered |
Underflow in Floating Point Numbers
Underflow is a condition in which a floating point number is too small to be represented in the given format. When a floating point number is too small, its magnitude is less than the minimum value that can be represented by its exponent. This condition is indicated by a special bit pattern in the exponent field of the floating point number's representation.
MSB of Mantissa with O
If a floating point number has an O in the MSB (most significant bit) of mantissa, it means that the number is very small and it cannot be represented accurately. In other words, the number is so small that it underflows the range of values that can be represented by the given format.
Overflow and Underflow
Overflow and underflow are two common problems that can occur in floating point calculations. Overflow occurs when a floating point number is too large to be represented in the given format. Underflow, on the other hand, occurs when a floating point number is too small to be represented in the given format.
Importance of Underflow
Underflow is an important issue in floating point calculations because it can lead to loss of precision and accuracy in the result. When a floating point number underflows, it is usually rounded to zero or the smallest representable value, which can result in significant errors in the final result.
Conclusion
In summary, a floating point number that has an O in the MSB of mantissa is said to have underflowed the range of values that can be represented by the given format. Underflow is an important issue in floating point calculations because it can lead to loss of precision and accuracy in the result.
Underflow is a condition in which a floating point number is too small to be represented in the given format. When a floating point number is too small, its magnitude is less than the minimum value that can be represented by its exponent. This condition is indicated by a special bit pattern in the exponent field of the floating point number's representation.
MSB of Mantissa with O
If a floating point number has an O in the MSB (most significant bit) of mantissa, it means that the number is very small and it cannot be represented accurately. In other words, the number is so small that it underflows the range of values that can be represented by the given format.
Overflow and Underflow
Overflow and underflow are two common problems that can occur in floating point calculations. Overflow occurs when a floating point number is too large to be represented in the given format. Underflow, on the other hand, occurs when a floating point number is too small to be represented in the given format.
Importance of Underflow
Underflow is an important issue in floating point calculations because it can lead to loss of precision and accuracy in the result. When a floating point number underflows, it is usually rounded to zero or the smallest representable value, which can result in significant errors in the final result.
Conclusion
In summary, a floating point number that has an O in the MSB of mantissa is said to have underflowed the range of values that can be represented by the given format. Underflow is an important issue in floating point calculations because it can lead to loss of precision and accuracy in the result.
A self relocating program is one which- a)Cannot be made to execute in any area of storage other than the one designated for it at the time of its coding or translation.
- b)Consists of a program and relevant information for its relocation.
- c)Can itself perform the relocation of its address- sensitive portions.
- d)All of the above
Correct answer is option 'C'. Can you explain this answer?
A self relocating program is one which
a)
Cannot be made to execute in any area of storage other than the one designated for it at the time of its coding or translation.
b)
Consists of a program and relevant information for its relocation.
c)
Can itself perform the relocation of its address- sensitive portions.
d)
All of the above
|
Harshad Mukherjee answered |
The self relocating program is one which can itself perform the relocation of its address sensitive portion. Code-optimization and target code generation comes under the pass-2 of the compiler and is machine independent.
Hash tables can contribute to an efficient average case solution for
- a)Counting distinct values
- b)Dynamic dictionary
- c)Range search
- d)Symbol table lookup
Correct answer is option 'D'. Can you explain this answer?
Hash tables can contribute to an efficient average case solution for
a)
Counting distinct values
b)
Dynamic dictionary
c)
Range search
d)
Symbol table lookup
|
|
Advait Shah answered |
Hash table can contribute to an efficient average case solution for symbol table look up problem
Which of the following is not a regular grammars?
- a)S →∈, A→ aS| b
- b)A→ abB |a B
- c)A→ Ba | Bab
- d)A → aB | a, B → Ab | b
Correct answer is option 'D'. Can you explain this answer?
Which of the following is not a regular grammars?
a)
S →∈, A→ aS| b
b)
A→ abB |a B
c)
A→ Ba | Bab
d)
A → aB | a, B → Ab | b
|
|
Avantika Yadav answered |
Mixup notation of left linear and right linear grammar is not allowed in regular grammar.
Any Directed Acyclic Graph must have- a)Atleast one vertex with indegree 1 and atleast one vertex with outdegree 0.
- b)Atieast one vertex with indegree 0 or atleast one vertex with outdegree 0,
- c)Atleast one vertex with indegree 0 and atleast one vertex with outdegree 0,
- d)Atieast one vertex with indegree 0 and atleast one vertex with outdeg ree 1
Correct answer is option 'C'. Can you explain this answer?
Any Directed Acyclic Graph must have
a)
Atleast one vertex with indegree 1 and atleast one vertex with outdegree 0.
b)
Atieast one vertex with indegree 0 or atleast one vertex with outdegree 0,
c)
Atleast one vertex with indegree 0 and atleast one vertex with outdegree 0,
d)
Atieast one vertex with indegree 0 and atleast one vertex with outdeg ree 1
|
|
Subham Menon answered |
Simplest DAG:
Which is not true about syntax and semantic parts of a computer language- a)Syntax is generally checked by the programmer.
- b)Semantics is the responsibility of the programmer.
- c)Semantics is checked mechanically by a computer.
- d)Both (b) and (c) above
Correct answer is option 'D'. Can you explain this answer?
Which is not true about syntax and semantic parts of a computer language
a)
Syntax is generally checked by the programmer.
b)
Semantics is the responsibility of the programmer.
c)
Semantics is checked mechanically by a computer.
d)
Both (b) and (c) above
|
Bibek Choudhary answered |
Syntax analysis checks for the syntax of the source program while the semantic analysis, deals with checking of semantic (meaning) of the source program output by syntax analysis phase and input to semantic analysis phase.
Note: The syntax checking is responsibility of the programmer while semantic analys is is checked at the runtime.
Note: The syntax checking is responsibility of the programmer while semantic analys is is checked at the runtime.
A simple two-pass assembler does which of the following in the first pass?
1. It allocates space for the literals.
2. It computes the total length of the program.
3. It builds the symbol table for the symbols and their values.
4. It generates code for all the load and stores register instruction.- a)1 only
- b)1 and 2
- c)1, 2 and 3
- d)4 only
Correct answer is option 'C'. Can you explain this answer?
A simple two-pass assembler does which of the following in the first pass?
1. It allocates space for the literals.
2. It computes the total length of the program.
3. It builds the symbol table for the symbols and their values.
4. It generates code for all the load and stores register instruction.
1. It allocates space for the literals.
2. It computes the total length of the program.
3. It builds the symbol table for the symbols and their values.
4. It generates code for all the load and stores register instruction.
a)
1 only
b)
1 and 2
c)
1, 2 and 3
d)
4 only
|
|
Gargi Sarkar answered |
Simple two-pass assembler:
1. Allocates space for the literals.
2. Computers the total length of program (syntax analysis).
3. Builds the symbol table for the symbols and their values.
1. Allocates space for the literals.
2. Computers the total length of program (syntax analysis).
3. Builds the symbol table for the symbols and their values.
Which of the following statements is true?- a)SLR parser is more powerful than LALR.
- b)IALR parser is more powerful than Canonical LR parser.
- c)Canonical LR parser is more powerful than LALR parser.
- d)The parsers SLR, Canonical GR, and LALR have the same power.
Correct answer is option 'C'. Can you explain this answer?
Which of the following statements is true?
a)
SLR parser is more powerful than LALR.
b)
IALR parser is more powerful than Canonical LR parser.
c)
Canonical LR parser is more powerful than LALR parser.
d)
The parsers SLR, Canonical GR, and LALR have the same power.
|
|
Abhijeet Unni answered |
The canonical LR parser is the most powerful parser that can recognize more grammars than other parsers.
Consider the grammar rule E → E1 - E2 for arithmetic expressions. The code generated is targeted to a CPU having a single user register. The subtraction operation requires the first operand to be in the register. If E1 and E2 do not have any common subexpression, in order to get the shortest possible code- a)E1 should be evaluated first
- b)E2 should be evaluated first
- c)Evaluation of E1 and E2 should necessarily be interleaved
- d)Order of evaluation of E1 and E2 is of no consequence
Correct answer is option 'B'. Can you explain this answer?
Consider the grammar rule E → E1 - E2 for arithmetic expressions. The code generated is targeted to a CPU having a single user register. The subtraction operation requires the first operand to be in the register. If E1 and E2 do not have any common subexpression, in order to get the shortest possible code
a)
E1 should be evaluated first
b)
E2 should be evaluated first
c)
Evaluation of E1 and E2 should necessarily be interleaved
d)
Order of evaluation of E1 and E2 is of no consequence
|
|
Pritam Chatterjee answered |
To optimize the solution evaluate the expression E2. Then we can calculate E1 and finally E1 will be one of operands that will be in register and we can perform subtraction directly. But if we follow the opposite then we have to make move and store operations.
Generation of intermediate code based on an abstract machine model is useful in compilers because- a)It makes implementation of lexical analysis and syntax analysis easier.
- b)Syntax-directed translations can be written for intermediate code generation.
- c)It enhances the portability of the front end of thecomplier.
- d)It is not possible to generate code for real machines directly from high level language programs.
Correct answer is option 'A'. Can you explain this answer?
Generation of intermediate code based on an abstract machine model is useful in compilers because
a)
It makes implementation of lexical analysis and syntax analysis easier.
b)
Syntax-directed translations can be written for intermediate code generation.
c)
It enhances the portability of the front end of thecomplier.
d)
It is not possible to generate code for real machines directly from high level language programs.
|
Shounak Sharma answered |
Generation of intermediate code based on an abstract machine model is useful in compilers because it makes implementation of lexical analysis and syntax analysis easier
A pictorial representation of the value computed by each statement in the basic block is- a)Tree
- b)DAG
- c)Graph
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
A pictorial representation of the value computed by each statement in the basic block is
a)
Tree
b)
DAG
c)
Graph
d)
None of these
|
|
Krithika Kaur answered |
A pictorial representation of the value computed by. each statement in the basic block is control flow graph.
Consider the following procedure declaration:
procedure p ;
x : integer;
procedure q;
begin x = x + 1 end;
procedure r;
x : integer;
begin x := 1; q ; write(x) end;
begin
x = 2;
r;
end;The output produced by calling P ina language with static scope will be- a)I
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
Consider the following procedure declaration:
procedure p ;
x : integer;
procedure q;
begin x = x + 1 end;
procedure r;
x : integer;
begin x := 1; q ; write(x) end;
begin
x = 2;
r;
end;
procedure p ;
x : integer;
procedure q;
begin x = x + 1 end;
procedure r;
x : integer;
begin x := 1; q ; write(x) end;
begin
x = 2;
r;
end;
The output produced by calling P ina language with static scope will be
a)
I
b)
2
c)
3
d)
4
|
Shounak Sengupta answered |
Since value output by program is the value of x which is declared in side DC function, so, output is 1 using static scoping.
Consider the following statements:S1 Static allocation bindings do not change at runtime.
S2: Heap allocation allocates and de-allocates storage at run time.
Which of the above statements is/are true- a)S1 is true and S2 is false
- b)S2 is true and S1 is false
- c)Both are true
- d)Both are fals
Correct answer is option 'C'. Can you explain this answer?
Consider the following statements:
S1 Static allocation bindings do not change at runtime.
S2: Heap allocation allocates and de-allocates storage at run time.
Which of the above statements is/are true
S2: Heap allocation allocates and de-allocates storage at run time.
Which of the above statements is/are true
a)
S1 is true and S2 is false
b)
S2 is true and S1 is false
c)
Both are true
d)
Both are fals
|
|
Pallabi Sharma answered |
- Static allocation binding do not change at runtime.
- Heap allocation allocated and deallocate storage at run time
Which one of the following statements is TRUE?- a)LR(1) parsing is sufficient for deterministic context-free languages
- b)Data flow analysis is necessary for run-time memory management.
- c)Symbol table is accessed only during the lexical analysis phase
- d)The LALR(1) parser for a grammar G cannot have reduce-reduce conflict if the LR(1) parser for G does not have reduce-reduce conflict.
Correct answer is option 'A'. Can you explain this answer?
Which one of the following statements is TRUE?
a)
LR(1) parsing is sufficient for deterministic context-free languages
b)
Data flow analysis is necessary for run-time memory management.
c)
Symbol table is accessed only during the lexical analysis phase
d)
The LALR(1) parser for a grammar G cannot have reduce-reduce conflict if the LR(1) parser for G does not have reduce-reduce conflict.
|
Varun Khanna answered |
Understanding LR(1) Parsing
LR(1) parsing is a type of bottom-up parsing technique widely used in compiler design. It can parse a larger class of deterministic context-free languages than many other parsing techniques.
Why is Option A True?
- Sufficient for Deterministic Context-Free Languages:
- LR(1) parsers can handle all deterministic context-free languages, meaning they can uniquely parse strings based on grammar with a single lookahead token.
- The "1" in LR(1) indicates that the parser uses one lookahead token to make parsing decisions, making it versatile and powerful.
Analyzing Other Options
- Option B - Data Flow Analysis:
- While useful for optimizing compilers, it is not necessary for run-time memory management. Memory management often involves allocation and deallocation rather than data flow analysis.
- Option C - Symbol Table Access:
- Symbol tables are utilized during various stages of compilation, including syntax analysis and semantic analysis, not just lexical analysis.
- Option D - LALR(1) Parser:
- LALR(1) parsers can have reduce-reduce conflicts even if the LR(1) parser does not. This is because LALR(1) combines states that can lead to ambiguities not present in the LR(1) version.
Conclusion
The correct answer is Option A, as LR(1) parsing is indeed sufficient for all deterministic context-free languages, making it a robust choice for parsing in compiler design.
LR(1) parsing is a type of bottom-up parsing technique widely used in compiler design. It can parse a larger class of deterministic context-free languages than many other parsing techniques.
Why is Option A True?
- Sufficient for Deterministic Context-Free Languages:
- LR(1) parsers can handle all deterministic context-free languages, meaning they can uniquely parse strings based on grammar with a single lookahead token.
- The "1" in LR(1) indicates that the parser uses one lookahead token to make parsing decisions, making it versatile and powerful.
Analyzing Other Options
- Option B - Data Flow Analysis:
- While useful for optimizing compilers, it is not necessary for run-time memory management. Memory management often involves allocation and deallocation rather than data flow analysis.
- Option C - Symbol Table Access:
- Symbol tables are utilized during various stages of compilation, including syntax analysis and semantic analysis, not just lexical analysis.
- Option D - LALR(1) Parser:
- LALR(1) parsers can have reduce-reduce conflicts even if the LR(1) parser does not. This is because LALR(1) combines states that can lead to ambiguities not present in the LR(1) version.
Conclusion
The correct answer is Option A, as LR(1) parsing is indeed sufficient for all deterministic context-free languages, making it a robust choice for parsing in compiler design.
Given two DFA’s M1, and M2. They are equivalent if- a)M1, and M2 has the same number of states
- b)M1 and M2 has the same number of final states
- c)M1, and M2 accepts the same language, i.e. L(M1 ) = L(M2)
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
Given two DFA’s M1, and M2. They are equivalent if
a)
M1, and M2 has the same number of states
b)
M1 and M2 has the same number of final states
c)
M1, and M2 accepts the same language, i.e. L(M1 ) = L(M2)
d)
None of the above
|
|
Ayush Basu answered |
Given two DFA’s M1 and M2. They are equivalent if the language generated by them is same i,e, L(M1) = L(M2)
Generation of intermediate code based on an abstract machine model is useful in compilers because- a)It makes implementation of lexical analysis and syntax analysis easier.
- b)Syntax-directed translations can be written for intermediate code generation.
- c)It enhances the portability of the front end of the compiler.
- d)It is not possible to generate code for real machines directly from high level language programs
Correct answer is option 'C'. Can you explain this answer?
Generation of intermediate code based on an abstract machine model is useful in compilers because
a)
It makes implementation of lexical analysis and syntax analysis easier.
b)
Syntax-directed translations can be written for intermediate code generation.
c)
It enhances the portability of the front end of the compiler.
d)
It is not possible to generate code for real machines directly from high level language programs
|
Shubham Chawla answered |
Explanation:
Intermediate code is code that is generated by a compiler between the source code and the target code. It is an abstraction of the target machine's code and is used to make it easier to translate the source code into machine code. The intermediate code is generally optimized and platform-independent, which makes it easier to port the compiler front end to different platforms.
The use of an abstract machine model in the generation of intermediate code has several advantages:
1. Easier implementation of lexical analysis and syntax analysis: The abstract machine model provides a simplified representation of the target machine's instruction set, making the implementation of the lexical and syntax analysis stages of the compiler easier.
2. Syntax-directed translations: Intermediate code generation can be done using syntax-directed translation techniques, which facilitates the construction of compilers.
3. Enhanced portability: The use of intermediate code makes it easier to port the front end of the compiler to different platforms. Since the intermediate code is platform-independent, only the back end of the compiler needs to be modified to generate code for different target machines.
4. Direct generation of code for real machines: Although it is possible to generate code for real machines directly from high-level language programs, the use of an abstract machine model provides a layer of abstraction that makes the process easier and more efficient.
Therefore, option C is the correct answer.
Intermediate code is code that is generated by a compiler between the source code and the target code. It is an abstraction of the target machine's code and is used to make it easier to translate the source code into machine code. The intermediate code is generally optimized and platform-independent, which makes it easier to port the compiler front end to different platforms.
The use of an abstract machine model in the generation of intermediate code has several advantages:
1. Easier implementation of lexical analysis and syntax analysis: The abstract machine model provides a simplified representation of the target machine's instruction set, making the implementation of the lexical and syntax analysis stages of the compiler easier.
2. Syntax-directed translations: Intermediate code generation can be done using syntax-directed translation techniques, which facilitates the construction of compilers.
3. Enhanced portability: The use of intermediate code makes it easier to port the front end of the compiler to different platforms. Since the intermediate code is platform-independent, only the back end of the compiler needs to be modified to generate code for different target machines.
4. Direct generation of code for real machines: Although it is possible to generate code for real machines directly from high-level language programs, the use of an abstract machine model provides a layer of abstraction that makes the process easier and more efficient.
Therefore, option C is the correct answer.
Operator-precedence parsing method is a parsing method. Which of the following statement is false about it?
1. It is bottom-up parsing method.
2. It must contains e-production.
3. It doesn’t contains two adjacent nonterminal symbols.- a)1 only
- b)2 only
- c)3 only
- d)1 and 3 only
Correct answer is option 'B'. Can you explain this answer?
Operator-precedence parsing method is a parsing method. Which of the following statement is false about it?
1. It is bottom-up parsing method.
2. It must contains e-production.
3. It doesn’t contains two adjacent nonterminal symbols.
1. It is bottom-up parsing method.
2. It must contains e-production.
3. It doesn’t contains two adjacent nonterminal symbols.
a)
1 only
b)
2 only
c)
3 only
d)
1 and 3 only
|
|
Aarav Malik answered |
Rules of operator precedence grammar is given as:
1. It must not contain any null-production.
2. It must not contain two adjacent non-terminal symbols.
1. It must not contain any null-production.
2. It must not contain two adjacent non-terminal symbols.
A group of bits that tell the computer to perform a specific operation is known as________- a)Instruction code
- b)Micro-operation
- c)Accumulator
- d)Register
Correct answer is option 'A'. Can you explain this answer?
A group of bits that tell the computer to perform a specific operation is known as________
a)
Instruction code
b)
Micro-operation
c)
Accumulator
d)
Register
|
Keerthana Sarkar answered |
Instruction code is the set of specific tasks to be performed
An LALR(1) parser for a grammar G can have shift-reduce (S-R) conflicts if and only if- a)The SLR(1) parser for G has S-R conflicts.
- b)The LR(1) parser for G has S-R conflicts
- c)The LR(0) parser for G has S-R conflicts.
- d)The LALR(1) parser for G has reduce-reduce conflicts.
Correct answer is option 'B'. Can you explain this answer?
An LALR(1) parser for a grammar G can have shift-reduce (S-R) conflicts if and only if
a)
The SLR(1) parser for G has S-R conflicts.
b)
The LR(1) parser for G has S-R conflicts
c)
The LR(0) parser for G has S-R conflicts.
d)
The LALR(1) parser for G has reduce-reduce conflicts.
|
|
Snehal Desai answered |
LALR(1) parser for a grammar, G can have shift- reduce ( S - R) conflicts if and only if there is a S - R conflict in its immediate sub-parsers (e.g. LR(1)). If LR(0) has S - R conflicts then they may get removed in LR (1), hence there will be no S - R conflict in LALR(1).
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