Translating an array reference a[i] into three address code requires
Answer: A
The address is the base plus the index times the element width. Multi dimensional arrays extend the same idea, with one multiplication per dimension in row major order. The multiplications by constant bounds are usually strength reduced afterwards.
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An array A of 4 byte elements is declared with lower bound 0. What offset from the base does A[7] occupy?
Answer: B
With a lower bound of zero the offset is simply the index times the element width: \[ \text{offset} = i \times w = 7 \times 4 = 28 \ \text{bytes} \] A non-zero lower bound \( l \) would require the base to be adjusted, giving \( (i - l) \times w \) instead. Note the offset is measured from the base, so \( \mathrm{A}[7] \) is the eighth element yet sits 28 bytes along, not 32.
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A two dimensional array B of 4 byte elements has 10 columns and lower bounds of zero. In row major order, what offset does B[2][3] occupy?
Answer: C
Row-major order lays each row out contiguously, so the linear index counts whole rows first: \[ \text{index} = i \times n_{cols} + j = 2 \times 10 + 3 = 23 \] \[ \text{offset} = 23 \times 4 = 92 \ \text{bytes} \] Column-major order would compute \( j \times n_{rows} + i \) instead, which is why the same subscript addresses different memory in Fortran and in C.
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Row major order stores the elements of a two dimensional array
Answer: D
The last subscript varies fastest in row major order. C and its relatives use it while Fortran uses column major, which matters when arrays are passed between languages. Mixing the two conventions silently transposes the data.
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Address computation for arrays is usually rearranged so that
Answer: A
Splitting the address into a constant part and an index dependent part lets the constant be folded at compile time. Only the varying portion then needs computing at run time. For a loop over an array the saving is repeated on every iteration.
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Translating a boolean expression by the value method produces code that
Answer: B
The value approach is needed when the result is stored or passed. It is less efficient than jumping code inside a condition, where the value is tested and immediately discarded. Compilers choose between the two according to how the result is used.
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Translating a boolean expression by the jumping method uses
Answer: C
Correct answer: inherited attributes naming the true and false destinations
The jumping translation never computes a boolean value at all: it passes down, as inherited attributes, the labels to jump to if the expression proves true and if it proves false. Each subexpression generates jumps to those, which is what gives the short-circuit evaluation of and and or for nothing.
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Assertion (A): Short circuit evaluation of a boolean expression avoids computing a value at all.
Reason (R): Each subexpression is given the destinations to jump to, so control reaches the right place without any result being stored.
Answer: D
Both are true and the reason gives the mechanism, and it also explains the semantics. Because the second operand is never reached when the first settles the outcome, side effects in it do not occur, which languages specify deliberately.
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Short circuit evaluation of a and b means that b
Answer: A
Once a is false the whole conjunction is false. Skipping b saves work and, more importantly, avoids any side effects the language guarantees will not happen. A call or an assignment hidden in the second operand simply never runs.
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Backpatching exists because
Answer: B
The destination of a forward jump is not yet known at the moment the jump must be written out. Leaving the field blank and noting where the instruction sits allows the value to be supplied later, which is what makes single pass translation of control constructs possible at all.
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