You know right shift 1 bit means dividing an integer by 2 and You know left shif

Question

You know right shift 1 bit means dividing an integer by 2 and You know left shift 1 bit means multiplying by 2.

(a)   Does this mean right shift is division exactly by 2? Explain this by shifting 0x9A 1 bit, 2 bits and 3 bits and show what values are 0x9A >> 1, 0x9A >> 2 and 0x9A >> 3.

(b)  Is it true that right shifting will always generate an integer of smaller or the same magnitude? When will n >> 1 be the same as n?

(c)   Given a general nonzero integer n with 0 < n < 255, what is / compute the minimum number of right shifts r to cause n >> r to become 0 (so this means n >> (r - 1) is still nonzero. What is that value?).

(d)   You have learned 0x6 << 4 = 0x60. Calculate 0x6 << n for n = 1 up to 8.

(e)  Can you conclude that left shift always create a bigger integer (of twice as big)? Why?

(f)   When do left shift 0x6 << n generate strange results? (there may be several n in (a) that this happens). Explain why you get odd results for such n.

Explanation / Answer

a) Not Always. If the number is even, its exactly divisble. If not, it gives (n-1)/2 , where n is the given number.

0x9A(154 in int) >> 1 =0x4D (77 in int)

0x9A >> 2 = 0x26 (38 in int) = (77 - 1)/2

0x9A >> 3 = 0x13 (19 in int)

b) Answer is no. If n=0, n>>1 is also 0.

c)Answer is floor(log n) + 1 where log is calculated with base 2.

d)

0x6 << 1 = 0xc = 12

0x6 << 2 = 0x18 = 24

0x6 << 3 = 0x30 = 48

0x6 << 4 = 0x60 = 96

0x6 << 5 = 0xc0 = 192

0x6 << 6 = 0x180 = 384

0x6 << 7 = 0x300 = 768

0x6 << 8 = 0x600 = 1536

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