What is the maximum number of bytes required for an instruction which uses Inherent addressing?
What is the maximum number of bytes required for an instruction which uses Inherent addressing?
Homework Answers
Inherent addressing also called implicit addressing mode. And for implicit addressing mode generally operend is hidden and mostly use accumulator is which is of 1 byte or 8 bit. So maximum byte required is 1 byte.
Example of inherent addressing is RAL, RLC, RRC etc
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What is the maximum number of bytes required for an instruction
which uses Inherent addressing?
Based on CIDR addressing scheme, A subnet is 128.12.0.17/29. What is the maximum number of unique...
Based on CIDR addressing scheme, A subnet is 128.12.0.17/29. What is the maximum number of unique IP addresses this subnet can have?
A BALR instruction appears in a program on line 00C. The instruction itself is 2 bytes...
A BALR instruction appears in a program on line 00C. The instruction itself is 2 bytes long. The next line contains an assembly USING directive. There is a variable called FIELD1 defined at line 2DA. (a) What is the displacement that will appear in an instruction which references FIELD1? (b) Assume the BALR instruction reads BALR R10,0. What does this instruction do? (c) What do we call R10? (d) In order to refer to the register as...
What instruction is needed to check weather the Upper bytes of AX and BX are equal?
What instruction is needed to check weather the Upper bytes of AX and BX are equal?
Which option is correct answer? QUESTION 18 4 points Save Answer A paging system, with 20-bit...
Which option is correct answer?
QUESTION 18 4 points Save Answer A paging system, with 20-bit addressing uses 512 bytes for the page size. How many pages are available in this paging scheme? 1024 0512 2048 4096 QUESTION 19 4 points Save Answer A paging system with 16-bit addressing is used with a page size of 512 bytes. What is the page number and offset, both in decimal, for the (Hex) address A01F ? Page 2016 and 1F16 offset into...
A byte-oriented stop and wait protocol uses frames of 143 bytes and acknowledgements of 4 bytes. ...
A byte-oriented stop and wait protocol uses frames of 143 bytes and acknowledgements of 4 bytes. The processing delays at the sender and receiver are 2.6 μs. The physical layer operates at 27 Mbit/s over a cable of length 14.0 km, on which signals travel at two-thirds the speed of light. What is the cycle time (the time from the start of transmission of a frame to the start of transmission of the next frame)? Give your answer in μs,...
Question 6: An IP packet of size 4576 bytes needs to be sent through a network. If the Maximum Tr...
I need the help step by step. Thank you for your help.
Question 6: An IP packet of size 4576 bytes needs to be sent through a network. If the Maximum Transport Unit (MTU) size allowed by the network is 1536 bytes, the packet will need to be fragmented. Considering that a minimum number of fragments should be used, for each fragment, find the values for: DF and MF flags Fragment offset field Length of each fragment Then answer the...
9 1. Which index register is used by the STOSD instruction? 2. In what way is...
9 1. Which index register is used by the STOSD instruction? 2. In what way is the CMPS instruction ambiguous? 3. Suppose a two-dimensional array of doublewords has three logical rows and four logical columns. Write an expression using ESI and EDI that address the third column in the second row. (Numbering for rows and columns starts at zero). 4. What is the maximum number of comparisons needed by the binary search algorithm when an array contains 1,024 elements?
A VM system uses segmentation with segment tables having maximum 256 entries and a maximum segment...
A VM system uses segmentation with segment tables having maximum 256 entries and a maximum segment size of 4096 bytes. A section of the segment table is shown below, with all numbers expressed in hexadecimal (with the H suffix): What is the physical address expressed in base 16 for a hexadecimal virtual address of 2DAC0H? D0FFH BEEFAC0H 43500H 399AFH 2DC9AFH C9AFH
What instruction is used to convert an integer into a double precision floating point number and...
What instruction is used to convert an integer into a double precision floating point number and stores it in the stack? a. FILD Ob.FLD OCFST Od. FSTP QUESTION 5 Which instruction copies a floating point number from the stack and does not pop the stack? a.FILD b.FLD OCFST Od.FSTP
Which instruction in R-type, I-type, and J-type instruction sets has least number of components in its...
Which instruction in R-type, I-type, and J-type instruction sets has least number of components in its data path?
Which option is correct answer?
QUESTION 18 4 points Save Answer A paging system, with 20-bit addressing uses 512 bytes for the page size. How many pages are available in this paging scheme? 1024 0512 2048 4096 QUESTION 19 4 points Save Answer A paging system with 16-bit addressing is used with a page size of 512 bytes. What is the page number and offset, both in decimal, for the (Hex) address A01F ? Page 2016 and 1F16 offset into...
I need the help step by step. Thank you for your help.
Question 6: An IP packet of size 4576 bytes needs to be sent through a network. If the Maximum Transport Unit (MTU) size allowed by the network is 1536 bytes, the packet will need to be fragmented. Considering that a minimum number of fragments should be used, for each fragment, find the values for: DF and MF flags Fragment offset field Length of each fragment Then answer the...
A VM system uses segmentation with segment tables having maximum 256 entries and a maximum segment size of 4096 bytes. A section of the segment table is shown below, with all numbers expressed in hexadecimal (with the H suffix): What is the physical address expressed in base 16 for a hexadecimal virtual address of 2DAC0H? D0FFH BEEFAC0H 43500H 399AFH 2DC9AFH C9AFH
What instruction is used to convert an integer into a double precision floating point number and stores it in the stack? a. FILD Ob.FLD OCFST Od. FSTP QUESTION 5 Which instruction copies a floating point number from the stack and does not pop the stack? a.FILD b.FLD OCFST Od.FSTP
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