Question

1.Write the "destination" register in the instruction 671A in a string of 4 bits.

2.The instruction 9158 uses two registers as operands, and a third register as a destination for the result.
Which registers are used for the operands?

		9 and 1
		1 and 5
		5 and 8
		9 and 8

3. Translate the following instruction into English: 54F2

		Add the bit patterns in registers F and 2 together as if they were presented in two's complement and leave the result in register 4.
		Add the bit patterns in registers F and 2 together as if they were presented in two's complement and leave the result in register 5.
		Add the bit patterns in registers F and 2 together as if they were presented in floating-point notation and leave the result in register 4.
		Add the bit patterns in registers F and 2 together as if they were presented in floating-point notation and leave the result in register 5.

4. Translate the following English phrase into a machine language instruction:

Place the hexadecimal value 34 into register 9.

5.Translate the following English phrase into a machine language instruction:

Place the byte value stored at memory cell 34 into register 9.

Answer 1

Answer is as follows :

1) As we have 671A where 6 set for instruction, 7 for destination register, 1 & A for source register.

So destination register is 7

2) In given Machine language 5 and 8 are two operand registers. Operand registers are also known as source register.

So option C is correct.

3) As we have instruction 54F2, which perform addition of contents present in register F and 2 and store the result in register 4.

All the operation done with integer values or they may get result with 2's complement. Destination register is 4 here.

So correct option is A

4) For placing the hexadecimal value 34 to register 9, we use the simple LOAD instruction with opcode 2. Opcode define the which instruction will be performed. And opcode is 2 here

So machine language instruction is 2934

Where 9 is destination register and 34 is data

5) For placing the value of memory cell 34 to register 9, we use the memory LOAD instruction with opcode 1. Opcode define the which instruction will be performed. And opcode is 1 here.

So machine language instruction is 1934

Where 9 is destination register and 34 is memory cell.

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