Symbol Table Management Questions and Answers

41.

A duplicate declaration in the same scope is detected by

A.looking the name up in the current scope before installing it
B.comparing the two declarations after the whole parse is done
C.the parser
D.the scanner

Answer: A

Correct answer: looking the name up in the current scope before installing it

Only the current scope is searched, because a declaration in an enclosing scope is legitimately shadowed rather than duplicated. That single lookup before installation is what distinguishes an error from ordinary nesting, and it is why the check must happen at declaration rather than afterwards.

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42.

Consider the following statements about symbol table scope handling.
1. A duplicate check must look only in the current scope.
2. A use lookup must follow the chain of enclosing scopes.
3. Both operations must search every scope in the program.

A.All of them
B.1 and 2 only
C.3 only
D.1 only

Answer: B

The two operations differ precisely in how far they search. Requiring the duplicate check to search outward would wrongly reject any legitimate shadowing declaration. Almost every block structured program relies on shadowing somewhere.

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43.

A symbol table entry's storage class attribute records

A.the hash bucket in which the entry has been placed
B.whichever source line carried the declaration of the name
C.whether the name is static, automatic or external
D.the number of bytes the name occupies in memory

Answer: C

The storage class determines where the variable lives and how long it persists. Code generation needs it to decide between a static address and a stack offset. The class also determines whether the variable survives between calls.

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44.

The offset recorded for a local variable is measured from

A.the beginning of the compiled object file
B.the first instruction of the whole program
C.the address of the previous local variable declared
D.the start of the activation record for its procedure

Answer: D

Locals live in the activation record created on each call, so their addresses are relative to a frame pointer. Only globals can be given an address fixed at compile time. Recursion is precisely why a local cannot have a fixed address.

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45.

A symbol table used by a compiler for a language with modules must handle

A.names exported by one module and imported by another
B.only the names declared within a single procedure
C.the machine registers available on the target
D.the sequence of passes the compiler will perform

Answer: A

Import and export lists determine which names cross module boundaries. The compiler records what it exports so a separate compilation of another module can check the uses. That record is what an interface or header file makes available.

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46.

Information a compiler writes for the linker about a module includes

A.a record of which optimisations were applied to the module
B.which names it defines and which it has left unresolved
C.the complete symbol table including all of the local names
D.the text of the module itself, kept for later reference

Answer: B

The linker needs exported definitions and imported references to connect modules. Local names are irrelevant to it, though they may be kept separately for a debugger. Keeping them in a separate section allows a binary to be stripped later.

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47.

Debugging information emitted alongside object code typically includes

A.the contents of the compiler's own hash table
B.a list of all the optimisations the compiler considered
C.variable names with their types and storage locations
D.the parse trees of every statement in the program

Answer: C

A debugger must map machine addresses back to source names. The information is derived from the symbol table but written in a separate standard format the debugger understands. The format must describe types, scopes and locations independently of the compiler.

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48.

Assertion (A): Symbol table information must be preserved in the object file for a debugger to be useful.
Reason (R): Without it the debugger has no way to relate a machine address to a source level name or type.

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

Answer: D

Both are true and the reason gives the mechanism, and it explains why stripped binaries are hard to debug. Optimisation complicates matters further, since a variable may live in different places at different points or nowhere at all.

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49.

Optimisation makes symbol table based debugging harder because

A.a variable may have no single storage location during its lifetime
B.the optimiser goes about deleting entries from the symbol table
C.optimised programs have no symbol table at all
D.the debugger cannot read an optimised object file

Answer: A

Correct answer: a variable may have no single storage location during its lifetime

Optimisation moves values between registers and memory, keeps them in different registers at different points, and may eliminate a variable entirely by folding or propagating it. A debugger asking where a variable is may find several answers or none, which is why debugging optimised code is difficult and why compilers emit location lists rather than single addresses.

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50.

A compiler that supports separate compilation must ensure that

A.all modules are compiled by the same invocation of the compiler
B.the declarations seen in each module agree with one another
C.every module is compiled in a single pass over its source
D.the symbol table of one module is embedded in the next

Answer: B

Header files or module interfaces supply the shared declarations. When two modules disagree, the mismatch may not surface until link time or, worse, until run time. Shared interface files are the usual way of preventing such disagreements.

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Symbol Table Management Questions and Answers

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