Memory-Reference Instructions - STA, LDA and BSA




Memory-Reference Instructions

In order to specify the microoperations needed for the execution of each instruction, it is necessary that the function that they are intended to perform be defined precisely. 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.

Table 5-4 lists the seven memory-reference instructions

The decoded D; for i = 0, 1, 2, 3, 4, 5, and 6 from the operation decoder that belongs to each instruction is included in the table. The effective address of the instruc­tion is in the address register AR and was placed there during timing signal T2 when I = 0, or during timing signal T3 when I = 1. The execution of the memory-reference instructions starts with timing signal T4• The symbolic de­scription of each instruction is specified in the table in terms of register transfer notation.

The actual execution of the instruction in the bus system will require a sequence of microoperations. This is because data stored in memory cannot be processed directly. The data must be read from memory to a register where they can be operated on with logic circuits. We now explain the operation of each instruction and list the control functions and microoperations needed for their execution.
A flowchart that summarizes all the microoperations is pre­sented at the end of this section

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Memory-Reference Instructions - STA, LDA and BSA

 

AND to AC

This is an instaruction that perform the AND logic operation on pairs of bits in AC and the memory word specified by the effective address. The result of

the operation is transferred to AC . The microoperations that execute this instruction are:
D0T4: DR <- M [AR]
D0T5: AC <- AC /\ DR, SC <--- 0

The control function for this instruction uses the operation decoder D0 since this output of the decoder is active when the instruction has an AND operation whose binary code value 000. Two timing signals are needed to execute the instaruction. The clock transition associared with  timing signal T4 transfers the operand from memory into DR . The clock transition associated with the next timing signal T5 transfers to AC the result of the AND logic operation between the contents of DR and AC. The same clock transition clears SC to 0, transfer­ring control to timing signal T0 to start a new instruction cycle.

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ADD to AC

 

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 Cout is transferred to the E (extended accumulator) flip-flop. The rnicroopera­tions needed to execute this instruction are
D1T4: DR← M[AR]
D1T5: AC← AC + DR, E← Cout , SC ← 0

 

Same Two timing signals, T, and T5, are used again but with operation decoder D1 instead of D0, which was used for the AND instruction. After the instruction is fetched from memory and decoded, only one output of the operation decoder will be active, and that output determines the sequence of microoperations that the control follows during the execution of a memory-ref­erence instruction.

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LDA: Load to AC

 

This instruction transfers the memory word specified by the effective address to AC . The microoperations needed to execute this instruction are

 

D2T4: DR← M [AR]
D2T5: AC← DR ,  ← 0

Looking back at the bus system shown in Fig. 5-4 we note that there is no direct path from the bus into AC . The adder and logic circuit receive information from DR which can be transferred into AC . Therefore, it is necessary to read the memory word into DR first and then transfer the content of DR into AC . The reason for not connecting the bus to the inputs of AC is the delay encountered in the adder and logic circuit. It is assumed that the time it takes to read from memory and transfer the word through the bus as well as the adder and logic circuit is more than the time of one clock cycle. By not connecting the bus to the inputs of AC we can maintain one clock cycle per microoperation.

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STA: Store AC

 

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:
D3T4: M [AR] ← AC, SC ← 0

 

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BUN: Branch Unconditionally


This instruction transfers the program to the instruction specified by the effective address. Remember that PC holds the address of the instruction to be read from memory in the next instruction cycle. PC is incremented at time T1 to prepare it for the address of the next instruction in the program sequence. The BUN instruction allows the programmer to specify an instruction out of sequence and we say that the program branches (or jumps) unconditionally. The instruction is executed with one microoperation:
D4T4: PC ← AR, SC ← 0

The effective address frpom AR is transferred through the common bus to PC .Resetting SC to 0 transfers control to T0• The next instruction is then fetched and executed from the memory address given by the new value in PC .

 

BSA: Branch and Save Return Address

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:

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M[AR] <-- PC, PC <-- AR + I

A numerical example that demonstrates how this instruction is used with a subroutine is shown in Fig. 5-10. The BSA instruction is assumed to be in memory at address 20. The I bit is 0 and the address part of the instruction has the binary equivalent of 135. After the fetch and decode phases, PC contains 21, which is the address of the next instruction in the program (referred to as the return address). AR holds the effective address 135. This is shown in part (a) of the figure. The BSA instruction performs the following numerical operation:

M[135] <-- 21, PC <-- 135 + 1 = 136

The result of this operation is shown in part (b) of the figure. The return address 21 is stored in memory location 135 and control continues with the subroutine program starting from address 136. The return to the original program (at address 21) is accomplished by means of an indirect BUN instruction placed at the end of the subroutine. When this instruction is executed, control goes to the indirect phase to read the effective address at location 135, where it finds the previously saved address 21. When the BUN instruction is executed, the effective address 21 is transferred to PC . The next instruction cycle finds PC with the value 21, so control continues to execute the instruction at the return address.

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ISZ: Increment and Skip if Zero

 

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.
Since it is not possible to increment a word inside the memory, it is necessary to read the word into DR, increment DR, and store the word back into memory. This is done with the following sequence of microoperations:
D6T4: DR <-- M [AR]
D6T5: DR <-- DR + 1
D,T,: M [AR] <-- DR,
if (DR = 0) then (PC ← PC + 1), SC ← 0

 

Control Flowchart

A flowchart showing all microoperations for the execution of the seven mem­ory-reference instructions is shown in Fig. 5-11. The control functions are indicated on top of each box. The microoperations that are performed during time T4, T5, or T, depend on the operation code value. This is indicated in the flowchart by six different paths, one of which the control takes after the instruction is decoded. The sequence counter SC is cleared to 0 with the last timing signal in each case. This causes a transfer of control to timing signal T0 to start the next instruction cycle.
Note that we need only seven timing signals to execute the longest instruction (ISZ). The computer can be designed with a 3-bit sequence counter. The reason for using a 4-bit counter for SC is to provide additional timing signals for other instructions that are presented in the problems section.

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Frequently Asked Questions

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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 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 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 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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Ans: The control gates for the seven flip-flops can be determined in a similar manner. For example, Table 5-6 shows that lEN may change as a result of the two instructions ION and !OF. view more..
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Ans: The 16-bit common bus shown in Fig. 5-4 is controlled by the selection inputs S2, S1, and 50• The decimal number shown with each bus input specifies the equivalent binary number that must be applied to the selection inputs in order to select the corresponding register. Table 5-7 specifies the binary numbers for S2S1S0 that select each register. Each binary number is associated with a Boolean variable x1 through x7, corresponding to the gate structure that must be active in order to select the register or memory for the bus. view more..




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