GATE CSE 1998 | Question: 1.17

Question

The octal representation of an integer is $(342)_8$. If this were to be treated as an eight-bit integer in an $8085$ based computer, its decimal equivalent is

• A. $226$
• B. $-98$
• C. $76$
• D. $-30$

Comments

Important line: If this were to be treated as an eight-bit integer in an 8085

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“The architects of the early integrated‑circuit‑based CPUs (Intel 8080, etc.) also chose to use two’s complement math... two’s complement technology was adopted in virtually all processors…” 
Since the 8085 is binary-compatible and an evolution of the 8080, it follows the same representation standard.

PROOF

NOTE: For almost all computing devices, the 2's complement system is used. 

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If a question mentions an "integer" in the context of a processor (like 8085, ARM, x86) and doesn't specify unsigned assume it's a signed integer represented in 2's complement.

Answer 1

$(3\; 4\; 2)_8 = (011 \; 100 \; 010)_2 = (11100010)_2.$

If we treat this as an 8 bit integer, the first bit becomes sign bit and since it is "1", number is negative. 8085 uses 2's complement representation for integers and hence the decimal equivalent will be $-(00011110)_2 = -30.$

Correct Answer: $D$

Comments

But where it is mentioned that numbers are stored n 2's complement representation ?

Is it by default ??

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given that, implemented in computer ===> 2's complement representation.

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what is the significance of specifically mentioning 8085 in the question...

Answer 2

First, write for each decimal equivalent binary code :

since 8 = 2³

write each digit in 3-bit binary 

(342)₈ = (011 100 010)₂  ignore initial zero 

(342)₈ = (226)₁₀ = (11100010)₂

since all processor use 2's complement number system(2's complement number system is weighted number system)

so 11100010 is a negative number

11100010 = 100010 = -2⁵ + 2 = -30

correct answer D

Comments

nice explanation