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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

1. $226$
2. $-98$
3. $76$
4. $-30$
edited | 1.9k views

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

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 $-(00011110)_2 = -30.$
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Sir, why haven't you considered 0 ( 011100010 ) as the sign bit? I calculated the answer as 226.

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they said its 8 bit integer
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Yes, I missed it. Thanks
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@arjun_sin, it is not mentioned anywhere in question, first bit will be signed bit, i am confused here when we consider 1st bit is signed bit and when we ignore 1st bit
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In this question they have mentioned 8bit integer in an 8085 therefore we are considering 8 bit from LSB to MSB, if nothing is mentioned then we have to consider it as a +ve number.
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can uh tell me meaning of 8085 based computer,
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Main thing is we have to take 8 bit from LSB to MSB

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@Arjun @ Arjun

Sir , why we are considering it as signed no?
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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.
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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

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nice explanation

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