Addressing modes refers to the way in which the operand of an instruction is specified.
The addressing mode specifies a rule of interpreting or modifying the address field of the instruction before the operand is actually executed.
When we address modes for 8086 instructions, they are divided in two categories: addressing modes for data addressing modes for branch.
The 8086 memory addressing modes provide flexible access to memory, allowing we to easily access variables, arrays, records, pointers and other complex data types. The key to good assembly language programming is the proper use of memory addressing modes.
A. 3 bits required
B. 6 bits are required
c. 8 bits are required
d. 15 bits are required
A. In the address mode selector, we need to specify 1 out of 7 addressing modes. Therefore, there must be 3 bits (2³ = 8)
B. There are 60 address registers. Therefore the bits to represent 60 numbers are 6 bits. (2^6 = 64)
C. As indicated, the memory bit word required is 8 bits.
D. Total Bits = Opcode + Address mode + Register Add + Memory ADD
32 bits = Opcode+ 3 bits + 6 bits + 8 bits
Opcode = 32 bits - 17 bits = 15 bits.
Therefore 15 bits are required for Opcode field.
Opcode = 3
AR = 20
a) Number of addressing modes = 4 = 22 , So it needs 2 bits for 4 values
Number of registers = 65 = 1000001 in binary , So it needs 7 bits
AR = 20
Bits left for opcode = 32 -(2+7+20) = 3
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i've had the same problem. i generally just ignore it, as a ton of these "verified answers" are completely wrong.