Question
1. The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with 4 fields: an opcode
1.
The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 7 addressing modes; a register address field to specify 1 of 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:
How large must the mode field be?
How large must the register field be?
How large must the address field be?
How large is the opcode field?
2.
Suppose we have the instruction Load 500. Given that memory and register R1 contain the values below:
Assuming R1 is implied in the indexed addressing mode, determine the actual value loaded into the accumulator and fill in the table below:
Mode | Value Loaded into AC |
---|---|
Immediate | |
Direct | |
Indirect | |
Indexed |
3.
Convert the following expressions from reverse Polish notation to infix notation.
W X Y Z + x
U V W X Y Z + x + x +
X Y Z + V W x Z + +
4.
Convert the following expressions from infix to reverse Polish (postfix) notation.
X x Y + W x Z + V x U
W x X + W x (U x V + Z)
(W x (X + Y x (U x V)))/(U x (X + Y))
5.
Show how the following values would be stored by machines with 32-bit words, using little endian and then big endian format. Assume each value starts at address 1016. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations.
456789A116
0000058A16
1414888816
Memory 100 600 R1 200 400 300 500 100 600 500 700 800Step by Step Solution
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