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

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

8 bit integer ===> all integers are 8 bits,

if it isn't signed, then we can't represent -ve integers, therefore it should be signed number.

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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.

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