Question

Write the hexadecimal notation for a MIPS machine language program that adds 42 to the value in register 2 placing the result in register 3 and then returns.

Example of running the program: Enter value for register 1: 2

Enter value for register 2: 3

Running MIPS program.

MIPS program completed normally.

$01 = 0x00000002 $02 = 0x00000003 $03 = 0x0000002D ...

Answer 1

Answer is as follows :

As we have to add the data value 42 with register 2 and store the result in register.

So accordingly register 2 = $02 and register 3 = $03

So MIPS instruction is :

ADDI $03, $02 , 42

So in this case

Instruction has 6 bit (31-26) opcode i.e. 001000

Source register $02 is of 5 bits (25-21) that represent as 00010 in binary

Destination Register $03 is of 5 (16-20) that represent as 00011 in binary.

And other 16 bits (15-0) represent data value i.e. 42 (in hexadecimal), so it is equal to 0000 0000 0100 0010 in 16 bit binary.

After joining all these we get

001000 00010 00011 0000000001000010

or we can write as

0010 0000 0100 0011 0000 0000 0100 0010

and this is equal to 20430042 in hexadecimal.

So Hexadecimal Notation for given scenario is 0x20430042

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