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Sonbull [250]
3 years ago
9

What is the size of the program counter for an avr that has a 1 kbyte rom capacity?

Computers and Technology
1 answer:
horsena [70]3 years ago
5 0

Answer:

9 bits

Explanation:

 We all recognize 1 K bytes as =1024 bytes.  

And the storage medium used for program instructions are the non-volatile flash memories. The MCUs are always 8 bits, and a piece of the instructions revenues two or one 16-bit words. Besides we know program memory magnitude through the device name, and as an instance, the ATmega64x line emanates with 64KB of flash, and the ATmega32x line emanates with 32 KB. And there is nobody like off-chip program memory, and all of the codes that are being implemented by the AVR essentially be on the on-chip flash. However, this restraint is not ever applied to the AT94 FPSLIC AVR/FPGA chips. Also, you must know that AVR is a family of microcontrollers established by ATEL in 1996. It is a Harvard architecture type single-chip 8-bit RISC microcontroller. And it is supposed to be the first microcontroller which applied flash memory for storing the program They are heavily used in embedded systems.  

1k byte = 1024 bytes, However, if the program instruction takes 16 bits.  

And hence it only wishes 512.  

Here and now 512 = 10^9, and henceforth you necessitate 9 bits at minimum.  

excluding you can somehow paginate that (such as through one-bit pagination, you will be necessitating only 8 bits and so on.)  

Therefore the answer here = 9 bits.

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) how many bits are used for host number on the child network (subnet) , b) how many usable addresses can exist on this child ne
vovikov84 [41]

Answer:

Machine’s IP=126.127.85.170, Machine’s Netmask=/27, Parent’s Netmask=255.255.240.0 .

a) Machine's Netmask = /27 : therefore no. of remaining bits for hosts =32-27 = 5 bits.

b) No. of usable addresses in the child network = 25 -2 = 32-2 =30 [Since first(network ID of the machine) and last ip (broadcast address of the machine ) addresses are not used ]

c) first usable address is on this child network (subnet) =

First, find out the network id of machine can be found out by doing bitwise AND machine's IP and Machine's subnet mask :

01111110. 01111111.01010101.10101010 (IP)

11111111. 11111111 .11111111 .11100000 (Subnet Mask)

01111110. 01111111. 01010101.10100000 (Network ID )

first usable address is on this child network (subnet) :   01111110. 01111111. 01010101.10100001

: 126.127.85.161

d) what the last usable address is on this child network (subnet) :  01111110. 01111111. 01010101.10111110

: 126.127.85.190

e) what the child network’s (subnet's) broadcast address is :

In the directed broadcast address , all the host bits are 1. Therefore, broadcast address :

01111110. 01111111. 01010101.10111111 (126.127.85.191)

f) what the child network's (subnet's) network number is : Network ID has already been calcuulated in part c .

01111110. 01111111. 01010101.10100000 (126.127.85.160)

6 0
3 years ago
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SashulF [63]

Answer:

A. Arbitrary code execution

Explanation:

A buffer-overrun, also known as buffer overflow occurs when a programmer makes an error during coding, which could make his system easily susceptible to attacks from a hacker. The root cause of this problem is an overflow of data in a buffer, and programs that are written in C and C+ + are more likely to be affected by this problem.  When a buffer-overrun occurs, it would most likely lead to arbitrary code execution.

Arbitrary code execution is an uncontrolled and unrestricted coding by the attacker which alters the original coding. The attacker could input more information into the coding, which alters the original information and which would make the application porous and accessible by just anyone. It is important that programmers run regular tests on their programs, so as to detect any coding errors.

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evablogger [386]

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3 years ago
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