Difficulty: Easy
Correct Answer: 8
Explanation:
Introduction / Context:Parallel interfaces transfer multiple bits at the same time across several signal lines. Classic microcomputer peripherals—especially Centronics-style printer ports—used an 8-bit parallel data path, reflecting the byte-oriented architecture of most systems.
Given Data / Assumptions:
Concept / Approach:
A “byte-wide” bus means 8 dedicated data lines (D0–D7) carry one full byte in a single strobe. Additional lines handle handshaking and status. Therefore, the number of simultaneous data bits equals eight.
Step-by-Step Solution:
Identify the physical interface: legacy printer port with D0–D7 data lines.Recognize that these eight lines convey a full byte per transfer cycle.Account for control lines (for example, STROBE, ACK) which do not carry data bits.Conclude that 8 bits are transmitted at once in typical microcomputer parallel transmission.Select option c (8).Verification / Alternative check:
Examine pinouts of Centronics/IEEE 1284 connectors which clearly denote eight data pins plus several control/status pins. Interface documentation confirms byte-wide operation.
Why Other Options Are Wrong:
a: 16-bit parallel buses exist (for example, certain internal buses) but are not the typical microcomputer parallel printer interface.
b: 9 is incorrect; confusion may arise from 9-pin serial connectors (DE-9 for RS-232).
d: 4-bit nibble modes are fallbacks for compatibility but not the standard full-speed parallel transfer mode.
e: Not applicable because 8 is standard.
Common Pitfalls:
Confusing serial connectors (DE-9) with parallel data width, or counting control lines as data lines. Also, IEEE 1284 has multiple modes, but the fundamental data path is byte-wide.
Final Answer:
8
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