Need help with interview question

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ManUtd_Vik

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How to implement i - 8 (8 bit length) ?

I answered that I would use a subtractor (modified adder) to construct carry skip or save or similar adder. He refused to accept it. Please help.

Thanks!
 

converting 8 to binary we get 1000. this means the lower 3 bits of the answer is going to be the lower 3 bits of i,no matter what. So you can instead use a 5 bit subtractor that does(i>>3) - 1 which is going to be a lot simpler and saves area, power and has higher speed. Whould this be a good answer?
 


if you mean by i>>3 shift right by three ... should work ... will try and get back to you

Thanks legendkiller
 

Example: I = 0xAA = b10101010 = 170

Shift 1 place right = 0x55 = b01010101 = 85
Shift 1 place right = 0x2A = b00101010 = 42
Shift 1 place right = 0x15 = b00010101 = 21
Subtract 1 = 0x14 = b00010100 = 20

Clearly subtracting 8 from 170 does not result in 20 so (i >> 3) - 1 does not work. It gives (i/8)-1 not 1-8.

The correct answer is to subtract 8 or add 0xF8, ignoring that a carry will be generated.

Brian.
 

@betwixt
actually the lower 3 bits of i are directly routed to the lower 3 bits of the answer hence you get 10100 from the 5 bit subtractor and 010 from the input i and concatinating it yields 10100010 which is 162. subtracting 8 is the actual answer but why use a full 8 bit subtractor when the lower 3 bits are going to depend on the lower 3 bits of i?

@ManUtd_Vik
you don't need a dedicated shifter circuit. the upper 5 bits can directly be roued to a 5 bit subtractor with the other input set to 1
 
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Although the lower three bits are the same, the answer is correct ONLY if the original value is >= 8. An 8-bit subtractor will indicate a 'borrow' occurred to show the number has turned negative, a 5 bit subtractor will not do that.

Brian.
 

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