GO Classes Test Series 2024 | Programming | Test 1 | Question: 6

Answer 1

Tmin = 100...000 and Tmax = 011...111.

Umin = 000…000 = 0 and Umax = 111...111.

bit pattern of 1 = 000...001.

In all above cases, each number is represented using 32 bits.

1. 1 < < 31 = 100...000 = Tmin, since we shifted bits left by 31 bits. 
2. 1 > > 31 = 000...000 = 0, since we shifted bits right by 31 bits.
3. ~ (1 > > 31)= ~ (000...000) = 111...111 = Umax ≠ Tmax, since we toggled each bit.
4. ! (1 > > 31) = ! (0) = ! (False) = True = 1.

Option A : True from point a.

Option B : False from point c.

Option C : False from point d.

Option D : True from point c.

Answer – A, D

Comments

shishir__roy

The minimum value in case of signed number should be all 1’s .

suppose for k=8bits

the minimum value for signed numbers will be 

1111 1111  =  – 127

https://en.wikipedia.org/wiki/Signed_number_representations#:~:text=The%20range%20of%20signed%20numbers,or%2011111111%20(%E2%88%920).

so for k=32 bits also the Tmin must be all 1’s right?

here ,  in case of k=8 bits 

if 1« 7

then it will give 

1000 0000 = -0

but it is greater than -127.

so by this OPTION A should be false.

---

@Pranavpurkar There are many representations of signed numbers in binary.

What you wrote is True for signed magnitude system, and what I wrote was for $2's$ complement system.

---

shishir__roy

But in question nothing about 2’s complement representation is mentioned. so , it depends on which representation we are considering , then also option a must be false. 

and signed numbers can be .

signed magnitude 

1’s complement 

2’s complement

---

@Pranavpurkar by default we have 2’s complement representation as signed number in C