A computer uses a memory unit of 512 K words of 32 bits each. A binary instruction code is stored in one word of the memory. The instruction has four parts: an addressing mode field to specify one of the two-addressing mode (direct and indirect), an operation code, a register code part to specify one of the 256 registers and an address part. How many bits are there in addressing mode part, opcode part, register code part and the address part?
1, 4, 8, 19
Two addressing modes require 1 bit; 256 registers require 8 bits.
\(512\times1024=2^{19}\)
The address field therefore needs 19 bits. The remaining opcode field has 32 − 1 − 8 − 19 = 4 bits. The required tuple is (1, 4, 8, 19).
Important Questions from Instruction Format - Teaching
In the assembly language of Mano’s Basic Computer, which of the following is not a pseudo-instruction (assembler directive)?
A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part?
A micro instruction format has micro operation field which is divided into 2 subfields F1 and F2, each having 15 distinct microoperations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field AD. The address space is of 128 memory words. The size of micro instruction is:
