The format of byte codes in EduVM are as follows:
The instructions are 32-bit and are represendted as hexadecimal instructions:
0x0A002006
0A is the opcode. The opcodes are 1 byte codes (in this case 0000 1010) 002 is the first argument. These are 12-bits. 006 is the second argument. These are also 12-bits.
The values of argument 1 and 2 work as follows:
000 -> 004 These represent the 5 registers the VM uses. Register A -> Register E
005 -> FFF represent integer literals. 0 -> 4095
So the instruction 0x0A002006 means: SET register C to 1 (6 - 5)
Explanation of the logic in vm_instruct_exec
word opcode = *instruction >> 24;
Here I extract the opcode from the instruction, this is easier seen if we convert the instruction to binary
0000 1010 0000 0000 0010 0000 0000 0110
|___________________________|
24-bits
If we right shift these bits out we'll be left with the opcode
word arg_b = *instruction & 2047
If we get the binary representation of 2047 we get
0000 0000 0000 0000 0000 1111 1111 1111 <-- This is 2047 in binary
0000 1010 0000 0000 0010 0000 0000 0110 <-- This is the instruction
If we perform a logical AND on the instruction with 2047 we're left with argument B
For arg_a we shift it right 12 bits to get argument A into the first 12 bits, then we logical AND it with 2047 to get its value
Any questions? Email me at jake.kgrogan@gmail.com