Question

Show the encoding of the following machine instructions and convert into hexadecimal.

1. STUR X7, [X9, #32]

2. SUBI X16, X4, #52

3. AND X5, X20, X3

Answer 1

Answer is as follows :

As we know that these instructions are 32 bit long. So machine instructions for given set are :

1) STUR X7, [X9, #32] : D-type Instruction

First 11 bits ( 31 to 21) are set for Opcode . So opcode is 0x7C0 for this instruction i.e. 11111000000.

Next 9 bits ( 20 to 12) are set for address or constants. So constant is #32 (in decimal) i.e. 000100000

Next 2 bits ( 11 to 10) are set to 00 for D- Type. So it is 00

Next 5 bits (9 to 5) are set for Address register i.e. X9 here. So it is 01001 in binary

Next 5 bits (4 to 0) are set for destination register i.e. X7 here. So it is 00111 in binary.

So we get

11111000000 000100000 00 01001 00111

or we can write as

1111 1000 0000 0010 0000 0001 0010 0111 i.e. 0xF8020127 in hexadecimal

So this instruction has

Machine Code - 1111 1000 0000 0010 0000 0001 0010 0111

Hexadecimal Code - 0xF8020127

2. SUBI X16, X4, #52 : I-type Instruction

First 11 bits ( 31 to 21) are set for Opcode . So opcode is 0x344 for this instruction i.e. 01101000100.

Next 11 bits ( 20 to 10) are set for address or constants. So constant is #52 (in decimal) i.e. 00000110100

Next 5 bits (9 to 5) are set for Source register i.e. X4 here. So it is 00100 in binary

Next 5 bits (4 to 0) are set for destination register i.e. X16 here. So it is 10000 in binary.

So we get

01101000100 00000110100 00100 10000

or we can write as

0110 1000 1000 0000 1101 0000 1001 0000 i.e. 0x6880D090 in hexadecimal

So this instruction has

Machine Code - 0110 1000 1000 0000 1101 0000 1001 0000

Hexadecimal Code - 0x6880D090

3. AND X5, X20, X3 : R-Type Instruction

First 11 bits ( 31 to 21) are set for Opcode . So opcode is 0x458 for this instruction i.e. 10001011000.

Next 5 bits (20 to 16) are set for Source register 2 i.e. X3 here. So it is 00011 in binary

Next 6 bits (15 to 10) are set for Shift OPerations. So this is arithmetic instruction so this will be 000000

Next 5 bits (9 to 5) are set for Source register 1 i.e. X20 here. So it is 10100 in binary

Next 5 bits (4 to 0) are set for destination register i.e. X5 here. So it is 00101 in binary.

So we get

10001011000 00011 000000 10100 00101

or we can write as

1000 1011 0000 0011 0000 0010 1000 0101 i.e. 0x8B030285 in hexadecimal

So this instruction has

Machine Code - 1000 1011 0000 0011 0000 0010 1000 0101

Hexadecimal Code - 0x8B030285