# Register-Reference Instructions

Register-reference instructions are recognized by the control when 07 = 1 and I = 0. These instructions use bits 0 through 11 of the instruction code to specify one of 12 instructions. These 12 bits are available in IR(0-11). They were also transferred to AR during time T2•

The control functions and microoperations for the register-reference instructions are. listed in Table 5-3. These instructions are executed with the clock transition associated with timing variable T3• Each control function needs the Boolean relation D7I'T3, which we designate for convenience by the symbol r. The control function is distinguished by one of the bits in IR(0-11). By assigning the symbol B, to bit i of IR, all control functions can be simply denoted by rB;. For example, the instruction CLA has the hexadecimal code 7800 (see Table 5-2), which gives the binary equivalent 011I 1000 0000 0000. The first bit is a zero and is equivalent to I'. The next three bits constitute the operation code and are recognized from decoder output D7• Bit 11 in IR is I and is recognized from 811• The control function that initiates the rnicrooperation for this instruction is D7I'T3B11 = rB11• The execution of a register-reference instruction is completed at time T3• The sequence counter SC is cleared to 0 and the control goes back to fetch the next instruction with timing signal T0•

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The first seven register-reference instructions perform clear, complement, circular shift, and increment rnicrooperations on the AC or E registers. The next four instructions cause a skip of the next instruction in sequence when a stated condition is satisfied. The skipping of the instruction is achieved by incrementing PC once again (in addition, it is being incremented during the fetch phase at time T1). The condition control statements must be recognized as part of the control conditions . The AC is positive when the sign bit in AC(IS) = 0; it is negative when AC(IS) = I. The content of AC is zero (AC = 0) if all the flip-flops of the register are zero. The HLT instruction clears a start-stop flip-flop S and stops the sequence counter from counting. To restore the operation of the computer, the start-stop flip-flop must be set manually.

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Ans: The timing signal that is active after the decoding is T3• During time T,, the control unit determines the type of instruction that was just read from memory. The flowchart of Fig. 5-9 presents an initial configuration for the instruction cycle and shows how the control determines the instruction type after the decoding. The three possible instruction types available in the basic computer are specified in Fig. 5-5. view more..
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Ans: Initially, the program counter PC is loaded with the address of the first instruction in the program. The sequence counter SC is cleared to 0, providing a decoded timing signal To. After each clock pulse, SC is incremented by one, so that the timing signals go through a sequence T0, T1, T2, and so on. The rnicrooperations for the fetch and decode phases can be specified by the following register transfer statements. view more..
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Ans: A program residing in the memory unit of the computer consists of a sequence of instructions. The program is executed in the computer by going through a cycle for each instruction. Each instruction cycle in turn is subdivided into a sequence of subcycles or phases. In the basic computer each instruction cycle consists of the following phases: view more..
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Ans: Register-reference instructions are recognized by the control when 07 = 1 and I = 0. These instructions use bits 0 through 11 of the instruction code to specify one of 12 instructions. These 12 bits are available in IR(0-11). They were also transferred to AR during time T2• view more..
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Ans: In order to specify the rnicrooperations needed for the execution of each instruction, it is necessary that the function that they are intended to perform be defined precisely. Looking back to Table 5-2, where the instructions are listed, we find that some instructions have an ambiguous description. This is because the explanation of an instruction in words is usually lengthy, and not enough space is available in the table for such a lengthy explanation. We will now show that the function of the memory-reference instructions can be defined precisely by means of register transfer notation. view more..
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Ans: This is an instruction that performs the AND logic operation on pairs of bits in AC and the memory word specified by the effective address. The result of view more..
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Ans: This instruction adds the content of the memory word specified by the effective address to the value of AC. The sum is transferred into AC and the output carry C,., is transferred to the E (extended accumulator) flip-flop. The rnicrooperations needed to execute this instruction are view more..
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Ans: This instruction transfers the memory word specified by the effective address to AC. The rnicrooperations needed to execute this instruction are view more..
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Ans: This instruction stores the content of AC into the memory word specified by the effective address. Since the output of AC is applied to the bus and the data input of memory is connected to the bus, we can execute this instruction with one microoperation: view more..
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Ans: This instruction is useful for branching to a portion of the program called a subroutine or procedure. When executed, the BSA instruction stores the address of the next instruction in sequence (which is available in PC) into a memory location specified by the effective address. The effective address plus one is then transferred to PC to serve as the address of the first instruction in the subroutine. This operation was specified in Table 5-4 with the following register transfer: view more..
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Ans: The BSA instruction performs the function usually referred to as a subroutine call. The indirect BUN instruction at the end of the subroutine performs the function referred to as a subroutine return. In most commercial computers, the return address associated with a subroutine is stored in either a processor view more..
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Ans: This instruction increments the word specified by the effective address, and if the incremented value is equal to 0, PC is incremented by 1. The programmer usually stores a negative number (in 2's complement) in the memory word. As this negative number is repeatedly incremented by one, it eventually reaches the value of zero. At that time PC is incremented by one in order to skip the next instruction in the program. view more..
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Ans: A computer can serve no useful purpose unless it communicates with the external environment. Instructions and data stored in memory must come from some input device. Computational results must be transmitted to the user through some output device. Commercial computers include many types of view more..
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Ans: The input register INPR consists of eight bits and holds an alphanumeric input information. The 1-bit input flag FGI is a control flip-flop. The flag bit is view more..
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Ans: The final flowchart of the instruction cycle, including the interrupt cycle for the basic computer, is shown in Fig. 5-15. The interrupt flip-flop R may be set at any time during the indirect or execute phases. Control returns to timing signal T0 after SC is cleared to 0. If R = 1, the computer goes through an interrupt cycle. If R = 0, the computer goes through an instruction cycle. view more..
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Ans: 1. A memory unit with 4096 words of 16 bits each 2. Nine registers: AR, PC, DR, AC, IR, TR, OUTR, INPR, and SC 3. Seven flip-flops: I, S, E, R, lEN, FGI, and FGO 4. Two decoders: a 3 x 8 operation decoder and a 4 x 16 timing decoder view more..
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Ans: he block diagram of the control logic gates is shown in Fig. 5-6. The inputs to this circuit come from the two decoders, the I flip-flop, and bits 0 through 11 of IR. The other inputs to the control logic are: AC bits 0 through 15 to check if AC = 0 and to detect the sign bit in AC( view more..
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Ans: The registers of the computer connected to a common bus system are shown in Fig. 5-4. The control inputs of the registers are LD (load), INR (increment), and CLR (clear). Suppose that we want to derive the gate structure associated with the control inputs of AR. We scan Table 5-6 to find all the statements that change the content of AR view more..

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