Question

How do you convert the following mips instructions into binary

li $t0, 7

li $t1, 2

please explain!

Answer 1

Actually load immediate is the mnemonic in MIPS and assembler translates this pseudoinstruction into the appropriate basic instruction.

Hence, li $t0, 7 will be translate into basic instruction as ori $t0, $zero, 7

and  li $t1, 2 will be translate into basic instruction as ori $t1, $zero, 2

Now MIPS binary code of ori $t0, $zero, 7 will be obtained as follows:-

6 bit opcode of ori which is 001101 followed by 5 bits code of source register $zero which will be 00000 followed by 5 bit opcode of register $t0 which is 01000 followed by 16 bits for immediate field which will be 0000 0000 0000 0111 for 7.

Hence binary code of li $t0, 7 will be

001101 00000 01000  0000 0000 0000 0111

Similarly ori $t1, $zero, 2 will have 6 bit opcode of ori which is 001101 followed by 5 bits code of source register $zero which will be 00000 followed by 5 bit opcode of register $t1 which is 01001 followed by 16 bits for immediate field which will be 0000 0000 0000 0010 for 2.

Hence binary code for ori $t1,$zero, 2 will be

001101 00000 01001 0000 0000 0000 0010

Please comment for any clarification