Final answer:
A table for 16 and 64 register systems changes the bit allocation for registers, thereby affecting the size of the Integer-Immediate field in a fixed-size instruction format. With 16 registers, 4 bits are used and with 64 registers, 6 bits are used for register identification, leaving more or fewer bits for the immediate field respectively.
Step-by-step explanation:
To create a table similar to the 32 register instruction format table for 16 and 64 registers, we need to consider that the number of bits for the register fields changes with the number of registers.
With 32 registers, usually 5 bits are required to uniquely identify each register (25 = 32). Consequently, for a system with 16 registers, we would need 4 bits (24 = 16), and for a system with 64 registers, we would need 6 bits (26 = 64).
Assuming an instruction format that has a fixed size, say 32 bits in total, the distribution of the bits among the different fields would change as the number of registers increases or decreases. Specifically, if fewer bits are used for the register fields, more bits remain available for the Integer-Immediate field, and vice versa.
For example, if an instruction consists of an opcode, two source register fields, a destination register field, and an immediate field, changing the number of registers affects the size of the immediate field:
- For 16 registers: 6 bits for opcode, 4+4+4 bits for the registers imply 14 bits are available for the Integer-Immediate field.
- For 32 registers: 6 bits for opcode, 5+5+5 bits for the registers imply 11 bits are available for the Integer-Immediate field.
- For 64 registers: 6 bits for opcode, 6+6+6 bits for the registers imply 8 bits are available for the Integer-Immediate field.