Question

The following data segment starts at memory address 2300h (hexadecimal) .data printString BYTE "Assembly is fun",0...

The following data segment starts at memory address 2300h (hexadecimal)

.data

printString BYTE "Assembly is fun",0

moreBytes BYTE 48 DUP(0)

dateIssued DWORD ?

dueDate DWORD ?

elapsedTime WORD ?

What is the hexadecimal address of dueDate?

2310h

2344h

2340h

2368h

Please explain the answer, I thought the answer was 2368h

Engineering   Computer-Science   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Suppose a program contains the data segment definitions val DWORD ? results BYTE 11 DUP(?), 0...
Suppose a program contains the data segment definitions val DWORD ? results BYTE 11 DUP(?), 0 and the code segment contains dtoa results, val Now, assuming that val contains FFFFFFE3, what bytes will be placed in memory at results by the dtoa macro? (Your answer should be in hex and should be the proper size)
Consider the following data segment .data array1 WORD 10 DUP(?),0,0 array2 DWORD 4 DUP(3 DUP(?)) What...
Consider the following data segment .data array1 WORD 10 DUP(?),0,0 array2 DWORD 4 DUP(3 DUP(?)) What decimal value the instruction : mov EBX , SIZEOF array2 stores in EBX. Show your work for full credit.
First clear all your general purpose registers by moving the value “0” into them. Initialize a...
First clear all your general purpose registers by moving the value “0” into them. Initialize a variable for a BYTE, WORD, DWORD storage size each with any desired value in the data segment. Initialize another variable called Result with the size of a DWORD and make the value as an uninitialized value. In the code segment, create a label called L1 that moves the variables in the appropriate sized register and making sure NOT to overwrite them in the process....
I am designing a custom PC and have RAM with a capacity 64GB. Assuming each byte...
I am designing a custom PC and have RAM with a capacity 64GB. Assuming each byte of data is accessible with discrete Memory Address, what is the required Address Bus Width to access all data within the RAM? Explain your answer.
Suppose a 1024-byte cache has an access time of 0.1 microseconds and the main memory stores...
Suppose a 1024-byte cache has an access time of 0.1 microseconds and the main memory stores 1 Mbytes with an access time of 1 microsecond. A referenced memory block that is not in the cache must be loaded into the cache and the reference started again. Answer the following questions: i. What is the number of bits needed to address the main memory? ii. If the cache hit ratio is 95%, what is the average access time for a memory...
Consider a memory system with a 16 bit address and the following address decode logic. The...
Consider a memory system with a 16 bit address and the following address decode logic. The signals enrom, enram, and endev are the device enable signals for ROM, RAM, and I/O devices, respectively. ◦ csrom <= '1' when ax(15 downto 12 = "1101" or ax(15 downto 13 = "111" else '0'; ◦ csram <= '1' when ax(15 downto 13) = "001" or ax(15 downto 12) = "0100" else '0'; ◦ csdev <= '1' when ax(15 downto 8) = "00001" else...
QUESTION 1 Convert the following code fragment to assembly code fragment, using instructions discussed in class....
QUESTION 1 Convert the following code fragment to assembly code fragment, using instructions discussed in class. if X > Y then Y=X else X=Y 2 QUESTION Suppose a user wants to do a system call. Assume that the service routine of this system call is at physical address 500. Suppose the user knows this address of the service routine. So instead of executing a system call, the user simply jumps to this location 500 (by executing “JMP 500”). Assume that...
(MIPS Assembly Language): Assume the following piece of MIPS code: label1: .text 0x4000fc lui $t0, 1022...
(MIPS Assembly Language): Assume the following piece of MIPS code: label1: .text 0x4000fc lui $t0, 1022 ori $t0, $t0, 2048 srl $t1, $t0, 18 sw $t0, 101($t1) slti $t2, $t1, 5 beq $t2, $0, label3 label2:... ... label3:... a) The code modifies a word in memory at the following address (circle one): (i) 1022 (ii) 101 (iii) 5 (iv) 10220000 (v) Other (provide your own answer): _________________ b) The following value is stored in memory (circle one): (i) 0x17 (ii)...
1. Assume there are three memory locations named h hand area, translate the following expression into...
1. Assume there are three memory locations named h hand area, translate the following expression into MIPS assembly language. No Floating point. Area= 1/2 b. h 2. Show the assembly statement that allocates a word in the data segment named area with an initial value of 10. 3. What are the contents of register $t0, $t1, and $t2? .data str:    .asciiz   "ab" .text main:   la $t0, str li    $t1, 0 loop: lb $t2, ($t0) beqz   $t2, eloop addi $t1,...
Assembly Programming in MIPS Write a program in MIPS assembler that performs the following computations: Prompts...
Assembly Programming in MIPS Write a program in MIPS assembler that performs the following computations: Prompts the user to enter two integers and stores the smaller one in the first data memory word (with the lowest address) and the larger one in the second data memory word. Then divides the larger word by the smaller one in integer and stores the integer quotient in the third data memory word and the remainder in the fourth data memory word. Finally, it...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT