GO Classes Test Series 2024 | Mock GATE | Test 12 | Question: 48

Answer 1

The $14$-bit unsigned offset allows you to encode a number that is as large as $2^{14}-1=16383$.

Answer 2

​​​​​Instruction start address => 5004 

Instruction size => 4B

PC storing the next instruction address as => 5008.

Max. unsigned number possible with 14 bits (unsigned offset) =                                                            2^14 -1 => 16383   

Target address = PC + offset = 5008 + 16383 => 21391

In the question, it is asked to find the number of instructions from the 
BEQ instruction (i.e. Number of instructions possible from instruction 5004)  

(21391 – 5008 + 1) / 4B  => 4096.

Here, (21391-5008+1) is used to calculate the number of addresses present starting from 5008 to 21391. 

Therefore, the answer will be 4096.

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Answer 3

Given:

14 bits are used to encode an "unsigned" offset

 Branch Instruction location:5004

PC value(while executing our branch instruction): 5008 (Note:any increment/decrement is always applied to this updated value )

so we can at Max jump: (2^14-1 )/ 4 =4095.75~ 4095 instruction away from "5008"(which is already one instruction away from our branch instruction ) 

so overall we can jump 4095+1=4096 instructions away from our branch instruction