Computer organization by zaky hamacher pdf

  1. Hamacher - Computer Organization (5th Ed)
  2. Computer Organisation : Fifth Edition - Carl Hamacher, Zvonko Vranesic, Safwat Zaky
  3. Computer Organisation : Fifth Edition - Carl Hamacher, Zvonko Vranesic, Safwat Zaky

Computer organization and embedded systems / Carl Hamacher. Safwat Zaky received his degree in Electrical Engineering and Home» Carl Hamacher» Computer Organisation» DOWNLOAD» edition» Fifth » FREE» PDF» Safwat Zaky» Zvonko Vranesic» Computer. Carl Hamacher et al.: Computer Organization, SE(ISBN ). Safwat Zaky received his degree in electrical engineering and in.

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Computer Organization By Zaky Hamacher Pdf

PDF. 2. Introduction - 1. PDF Chapter 8 [HAMACHER]. [HAMACHER]: Computer Organization by Carl Hamacher, Zvonko Vranesic, Safwat Zaky; 6th Edition. Additional Resources. Bluespec Resources. Bluespec Training Resources. Carl Hamacher, Zvonko Vranesic, Safwat Zaky: Computer Organization, 5th Edition, Tata Instruction set Architecture and 3) Computer Organization. 1. Computer organization V. Carl Hamacher, Zvonko G. Vranesic, Safwat G. Zaky. [electronic resource] - 3rd ed. New York McGraw-Hill - McGraw-Hill computer.

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Why download extra books when you can get all the homework help you need in one place? Computer Organization Fifth Edition Publisher: Ltd Publication Date: Softcover Book Condition: New Edition: About this title Synopsis: All books are new. We accept payments by the following methods: Credit Card: Visa, MasterCard, or American Express. Cheques may be made payable to BookVistas. Shipping Terms: Add to Wants. Previous 6 months Previous 1 month: MOV R5, 3 Same code as.

See the solution to Problem 2.

Register assignment: The record pointer is register R0, and registers R1, R2, and R3, are used to accumulate the three sums, as in Figure 2. Assume that the list is not empty. If the ID matches that of a later record, it will be inserted immediately after that record, including the case where the matching record is the Tail. Modify Figure 3. If the list is empty, the result is unpredictable because the second instruc- tion compares the new ID with the contents of memory location zero.

Replace Figure 3. One memory access is needed to fetch the instruction and 4 to execute it. L 9,D2 The number of bits shifted must be less than 8. B The destination operand must be a data register. Also the source operand is outside the range of signed values that can be represented in 8 bits.

W 83 5 5 0 after 2nd ADD. W 4 5 0 after 3rd ADD. W 3 5 0 after 4th ADD.

W 34 2 5 0 after 5th ADD. W 1 5 0 after last MOVE. We have assumed that the assembler uses short absolute addresses. L, etc.

Otherwise, 3 more words would be needed. Program 2 destroys the original list. A program for string matching: Therefore, the largest value of n that this program can handle is 14, because the largest number that can be stored in a byte is Assume that most of the time between successive characters being struck is spent in the two-instruction wait loop that starts at location READ.

Assume that register A4 is used as a memory pointer by the main program. B CR,D0 Check for end-of-line character. W 3, A0 Wait for character.

B A1 ,D0 Load character into D0. RTS Return. W 3, A2 Wait for display. B D0, A3 Send character to display. A stack structure like that shown in Figure 3. The main program uses register A0 as a memory pointer, and uses register D0 to hold the character read. L 16 A7 ,D0 Load long word containing character into D0.

B D0, A1 Send character to display. L 1,A1 Increment A1 Modulo k.

Hamacher - Computer Organization (5th Ed)

L A1,A2 Check if queue is full. B D0,[A5] Append byte. L A1,A2 Check if queue is empty. Using the same assumptions as in Problem 3. A program to reverse the order of bits in register D2: The trace table is: Assume the list address is passed to the subroutine in register A1.

When the subroutine is entered, the number of list entries needs to be loaded into D1. Because addresses must be incremented or decremented by 2 to handle word quantities, the address mode A1,D1 is no longer useful. Also, since the initial address points to the beginning of the list, we will scan the list forwards. Use D4 to keep track of the position of the largest element in the inner loop and D5 to record its value. We will use registers D1, D2, and D3 to accumulate the three sums.

Assume also that the list is not empty. Hence, the BNE instruction will test the correct values.

Computer Organisation : Fifth Edition - Carl Hamacher, Zvonko Vranesic, Safwat Zaky

In the program of Figure 3. Modify the program as follows. If the ID of the new record is less than that of the head, the program in Figure 3. If the list is not empty, the program continues until A2 points to the Tail record.

To correct behavior, modify the program as follows. Intel IA 3. Initial memory contents are: Only one operand can be in memory.

Scale factor can only be 1, 2, 4, or 8.

Computer Organisation : Fifth Edition - Carl Hamacher, Zvonko Vranesic, Safwat Zaky

An immediate operand can not be a destination. ESP cannot be used as an index register.

To change characters from lowercase to uppercase, change bit b5 from 1 to 0. Append routine: Remove routine: Assume that register ECX is used as a memory pointer by the main pro- gram. RET Return. The potential advantage is that the inner loop should execute faster. Otherwise, the same input data would be read a second time.

A subroutine is called by a program instruction to perform a function needed by the calling program. An interrupt-service routine is initiated by an event such as an input operation or a hardware error. Hence, it must not affect any of the data or status information relating to that program.


If execution of the interrupted instruction is to be completed after return from interrupt, a large amount of information needs to be saved. This includes the contents of any temporary registers, intermediate results, etc.

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An alternative is to abort the interrupted instruction and start its execution from the beginning after return from interrupt. In this case, the results of an instruction must not be stored in registers or memory locations until it is guaranteed that execution of the instruction will be completed without interruption. When an interrupt request is received from either A or B, interrupts from the other device will be automatically disabled until the request has been serviced.

However, interrupt requests from C will always be accepted. Interrupts are disabled before the interrupt-service routine is entered. Once de- vice turns off its interrupt request, interrupts may be safely enabled in the pro- cessor. If the interface circuit of device turns off its interrupt request when it receives the interrupt acknowledge signal, interrupts may be enabled at the be- ginning of the interrupt-service routine of device.

Otherwise, interrupts may be enabled only after the instruction that causes device to turn off its interrupt request has been executed. Yes, because other devices may keep the interrupt request line asserted.

Transfer of control among various programs takes place as shown in the diagram below. A binary variable, indicating whether a block is full and ready for processing. Number of characters read. Points at the location where the next input character is to be stored. Points at the location of the block to be processed by PROG. Two memory buffers are needed, each capable of storing a block of data.

Assume that the interface registers for each video terminal are the same as in Figure 4. Note that depending on the processor, several instructions may be needed to perform the function of one of the instructions used below. If several devices are ready at the same time, the routine will be entered several times in succession. In case a, POLL must be executed at least times per second.

The equivalent condition for case b can be obtained by considering the case when all 20 terminals become ready at the same time.

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