Symbol Table Management Questions and Answers

Q: Arrange the actions taken when a compiler enters and leaves a nested block.1. resolve uses within the block against the scope chain2. create or push a new scope on entering the block3. discard or pop the scope on leaving the block4. install the declarations found in the block into that scope

Correct answer: 2 - 4 - 1 - 3The scope must exist before anything can be installed in it, and the declarations must be installed before uses can be resolved against them. Discarding comes last, when the block ends and its names go out of scope. Resolving before installing would fail on any name declared in the block itself.

21.

Arrange the actions taken when a compiler enters and leaves a nested block.
1. resolve uses within the block against the scope chain
2. create or push a new scope on entering the block
3. discard or pop the scope on leaving the block
4. install the declarations found in the block into that scope

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

Answer: C

Correct answer: 2 - 4 - 1 - 3

The scope must exist before anything can be installed in it, and the declarations must be installed before uses can be resolved against them. Discarding comes last, when the block ends and its names go out of scope. Resolving before installing would fail on any name declared in the block itself.

Discuss in Forum

22.

An inner declaration that hides an outer one of the same name is said to

A.overload the outer declaration with an additional meaning
B.override the outer declaration permanently for the whole program
C.alias the outer declaration so both refer to one location
D.shadow the outer declaration for the extent of the inner block

Answer: D

Shadowing is temporary and lexical: the outer declaration becomes visible again on leaving the block. Overloading and aliasing are quite different relationships between names. Confusing the three is one of the commonest terminological errors in this area.

Discuss in Forum

23.

State whether the following is true or false.
A name declared in an inner block permanently removes the outer declaration from the symbol table.

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

Answer: A

The outer entry is hidden rather than deleted, and becomes visible again when the inner block ends. A stacked or chained table implements this restoration naturally. Deleting the outer entry would make the enclosing code fail to compile.

Discuss in Forum

24.

A language with dynamic scoping resolves a name using

A.the first declaration the compiler happens to find
B.the most recent binding in the call chain at run time
C.the innermost enclosing block in the program text
D.the declaration in the outermost scope of the program

Answer: B

Correct answer: the most recent binding in the call chain at run time

Dynamic scoping looks outward through the chain of active calls rather than outward through the enclosing text, so which declaration a name refers to depends on who called the procedure. It cannot be resolved at compile time, which is why static scoping is almost universal in modern languages.

Discuss in Forum

25.

Assertion (A): A compiler for a statically scoped language can resolve every name at compile time.
Reason (R): Visibility depends only on the nesting of blocks in the program text, which the compiler can see in full.

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

Answer: C

Both are true and the reason gives the mechanism, and it explains why static scoping is universal. A dynamically scoped language would have to search a chain of bindings at run time, costing both speed and predictability.

Discuss in Forum

26.

A name space in a language such as C separates

A.the names declared in different source files of a program
B.the names of types from the names of the values they hold
C.the identifiers used by the compiler from those used by the user
D.identifiers used as tags, labels, members and ordinary variables

Answer: D

The same spelling may be a struct tag and a variable simultaneously without conflict. The compiler keeps separate tables or tags each entry with the space it belongs to. Languages differ considerably in how many such spaces they define.

Discuss in Forum

27.

Overloading a name means that

A.several declarations share a name and are distinguished by signature
B.an inner declaration hides an outer one of the same name
C.the same storage location is reachable through two different names
D.a name gets used somewhere before it has ever been declared

Answer: A

Overload resolution selects among the candidates using the argument types at the call. The symbol table must therefore be able to hold several entries under one name. A chain of candidates under one key is the usual arrangement.

Discuss in Forum

28.

Resolving an overloaded call requires the symbol table to hold

A.no entries at all
B.all the candidate declarations reachable at that point
C.only the most recently declared of the candidates
D.a single entry with the parameter types omitted

Answer: B

Every visible candidate must be available so the best match can be chosen. Discarding all but one at declaration time would make correct resolution impossible. Some languages also permit the return type to participate in the choice.

Discuss in Forum

29.

A forward reference is a use of a name that

A.refers to a name in an enclosing procedure's scope
B.occurs inside a loop that has not yet been entered
C.appears in the text before the declaration of that name
D.refers to a name declared in a different source file

Answer: C

Correct answer: appears in the text before the declaration of that name

A forward reference forces the compiler either to make two passes, or to record the use and patch it when the declaration arrives. Languages requiring declaration before use avoid the difficulty; those permitting mutual recursion between procedures must deal with it somehow.

Discuss in Forum

30.

A language requiring declaration before use simplifies compilation because

A.the symbol table can then be implemented as a simple flat array
B.no scoping rules are needed for such a language at all
C.the parser can then be written without any lookahead
D.the attributes of every name are known when the use is compiled

Answer: D

Correct answer: the attributes of every name are known when the use is compiled

With the declaration guaranteed to have been seen already, the type, the scope and the storage are all in the symbol table when the use is compiled, so code and checks can be generated on the spot. Without that guarantee the compiler must defer, and a second pass or a patching mechanism becomes necessary.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Symbol Table Management Questions and Answers || Symbol Table Management MCQ Questions and Answers || Symbol Table Management GK Questions and Answers || Symbol Table Management GK MCQ Questions || Symbol Table Management Multiple Choice Questions and Answers || Symbol Table Management GK || GK on Symbol Table Management || Compiler Design Questions and Answers || Compiler Design MCQ Questions and Answers || Compiler Design GK Questions and Answers || GK on Compiler Design

Symbol Table Management Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications