Made easy CBT

Question

https://gateoverflow.in/?qa=blob&qa_blobid=11818269066433190011

Answer 1

5 bits required to specify 32 registers for register operand. So, 2 register operands 10 bits. 6 (⌈log₂49⌉) bits required for opcode(since 49 instructions). So (10 + 6 = ) 16 bits required. So left no. of bits = 24 - 16 = 8 to represent immediate operand.

As the operand is signed integer, we can represent -2⁷ to 2⁷ - 1. So, the max value will be -2⁷ = -128.

Comments

Isn't sign magnitude range is -(2^n-1 -1) to (2^n-1 -1)? So shouldn't answer be -127?

---

Thats exactly what my doubt was..!

Even i was getting -127

But finally researched that we represent -ve numbers as 2's compliment only in computers, hence -128

---

Then what's the point of signed magnitude? Should we always use 2's complment unless mentioned otherwise?

---

I feel whenever there is something related to a computer we have to go by the 2's compliment representation only

There's no clear explanation.... Even I was confused for a long time regarding this