|-------------------------------------------------16 bits-------------------------------------------------------|
Opcode ( 4 bits) |
Register (6 bits ) |
Register (6 bits ) |
so that Total no of opcode operation is =2^4=16 operation
so that there exist 12 2's address instruction which uses the register reference & 12 1-address memory reference instructions
then no of Free opcode operation after 2 address =16-12=4
then no of 12 1-address memory reference instructions
4
|---------------------------------------------16 bits ------------------------------------------------------------|
|-----------------6 bits --------------------------------|--------------------10 bits ---------------------------|
Memory 1 kb = 2^10 ( 10 bits for memory ) ( 16 -10=6 bits for opcode)
then no of opcode bits for 1 address for the memory reference (6-4 =2 bits )
((2^4-12)*2^2-12)*2^10 = 4096 O-address instructions are possible
how many =4096