The local declarations of a procedure are usually kept in
Answer: C
Nesting the tables mirrors the nesting of scopes in the source. Discarding or unlinking the local table on leaving the procedure restores the enclosing visibility. The table itself is often retained for the benefit of a debugger.
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The fields of a record type are usually kept in
Answer: D
Correct answer: a table of their own attached to the record's type entry
The fields form a scope of their own: two records may both have a field called next without conflict, and a field name means nothing outside its record. Giving each record type its own table expresses that directly, and the table also carries the offsets computed during declaration processing.
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A record field name and a variable name that coincide
Answer: A
The field is reached only through a record value, so no ambiguity arises. This separation is what allows a field called next in many different record types. Without it, field names would compete with ordinary variables for the same space.
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The offset of a field within a record is computed by
Answer: B
Each field is placed at the running total, which then advances by that field's width. Alignment requirements may insert padding, which the total must account for. Reordering the fields can reduce that padding considerably.
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A record has fields of widths 4, 8 and 2 bytes laid out in that order with no padding. What offset does the third field occupy?
Answer: C
Correct answer: 12
With no padding the offsets are cumulative from zero: the first field at 0, the second at 4, and the third at \( 4 + 8 = 12 \). The offset is the running total of the widths before the field, which is why the third field’s own width of 2 does not enter the calculation.
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A hash table with 128 buckets holds 512 entries using separate chaining. What is the average chain length?
Answer: D
Correct answer: 4
\[ \frac{512}{128} = 4. \] This is the load factor, and it is what governs lookup cost under separate chaining: each search examines about four entries on average. Doubling the number of buckets would halve it, which is why hash tables are grown as they fill.
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Enlarging a hash table when it becomes too full requires
Answer: A
The bucket a name belongs to depends on the table size, so every entry must be placed again. The cost is amortised over the many insertions between resizes. Doubling the size each time keeps the amortised cost per insertion constant.
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The symbol table must be consulted during type checking in order to
Answer: B
An expression's type is built from the types of its operands, and the types of names come from their declarations. Without the table no typing would be possible at all. Every operator then checks its operand types against what it will accept.
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An undeclared identifier is detected by
Answer: C
Correct answer: a failed lookup through the whole chain of enclosing scopes
The name is looked up in the innermost scope, then in each enclosing one in turn, and only when every scope has been searched can it be declared undeclared. The parser accepts the identifier as syntactically valid, so this is a semantic error rather than a syntactic one.
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Installing a placeholder entry for an undeclared name serves to
Answer: D
One diagnostic per undeclared name is far more useful to the programmer than one for every occurrence of it, which in a long procedure could run to dozens of identical messages. No code is generated for a program containing such an error, so the placeholder affects only the reporting rather than the translation.
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