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3 years ago

How does violating the SOLID principles make code hard to test?

Computers and Technology

1 answer:

Kamila [148]3 years ago

4 0

Answer:

SOLID is the acronym form of the first five principle of object oriented programming. It basically provide the high cohesive units and used in the testing purpose. The SOLID principle combined together to develop a software easily and it is easy to maintain.

By applying the SOLID principle it makes easy to understand it also provide the better quality in the software. Therefore, violating the SOLID principle it become difficult the code to test.

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Conduct a site research on a specific property using the concepts discussed during this week.

Anni [7]

Answer:Penis!

Explanation:

Haha! Enjoy your day! Hahahahaha!

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3 years ago

Outline the steps necessary to prepare a storage device for storing files. Identify and briefly describe the tools used to prepa

Fantom [35]

Answer:

Outline the steps necessary to prepare a storage device for storing files.

Link the storage unit to a network, whether it be a laptop computer or a mobile device.
Firewall it with sufficient anti-virus so that it also does not affect the linked system.
The program must auto-install the storage device on the network after quarantine, and authorize it.
Test the file that requires to be moved to the storage device, then upload it to the storage device after anti virus is safe.
Should not directly pull the storage device out of the network, select the exit device option correctly and only then remove the device after it has been accepted.

Identify and briefly describe the tools used to prepare a storage device for storing files.

EFSDump - Users can share encrypted files by carrying certificates by using EFSDump. It is usually very time consuming to audit the users who have privileges. But it is very simple to list users who have access to encrypted files using EFSDump.

SDelete - Suppose Windows operating system eliminates the index and prevents access when we delete a file. All the same, attackers can still recover the contents of the file. Files that are encrypted by EFS leave the file behind on the disk. Using the SDelete tool we can modify free space to prevent the recovery of deleted files.

Explain why file systems are critical components of an operating system. What is their purpose?

Hypothetically defining file system is nothing more than the way files are called and logically organized for storage.

Ex: DOS, Windows, and Unix-based operating systems all contain various file systems where data is organized in tree structure.

Usually defines the naming of files, how long the suffix can be. The file system often includes a format indicating the access to a file.

Specify the file system or file systems that would be available with that operating system.

A file system controls the mechanism by which data is stored and accessed on a disk. This deals mainly with internal disk operations such as naming files, records, directories, documentation, storage management, and other protocol accesses.

FAT (File Allocation Table)
NTFS (New Technology File System)

Describe from an administrator perspective what is necessary to prepare a storage device for access by a user.

Connection to the storage unit is equipment-oriented. The Windows Explorer summary contains optical drives, USB keys and hard drives as separate devices, making it much easier to navigate a new computer.

To connect a Linux computer you must mount the drive by connecting it to a directory.

Usually mounting a device is a manual operation, since only the administrator determines which directory the system attaches to.

Graphical interfaces are seldom seen on Linux servers. When you plug a storage device into a USB port on your computer, nothing can happen.

Usually the mount command looks like mount/dev / sdb1/mnt, where /dev / sdb1 is the system and /mnt is the directory you want to mount it.

List a few ways that a user can utilize a file system.

System output can be improved not only by choosing the most appropriate file system(s) but also by using the various choices available for most file systems.

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3 years ago

Lynn has created a quarterly sales report using a word processor. The document is confidential, and Lynn wants to secure it with

sweet-ann [11.9K]

Stop cheating

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2 years ago

Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations,

Fiesta28 [93]

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary denary or decimal

11111111 = 2⁸ = 256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s = /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11110000.000000 /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1  1  1 1 1 1

128 64 32 16 8 4 2 1

step 3

in the ip network address: 198.16.0.0/19 (subnet representation) we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts+16: 198.16.16.0 - 198.16.31.255

+16: 198.16.32.0- 198.16.47.255 etc

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11111000.000000 /21

now we have added 5 1s in the third field to reserve 11 0s

subnet mask: 255.255.8.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1 1 1  1 1 1

128 64 32 16 8  4 2 1

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (subnet representation) we increment this using 8

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts +16: 198.16.24.0- 198.16.31.255

+16: 198.16.32.0- 198.16.112.255 etc

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11110000.000000 /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1 1 1 1 1 1 1

128 64 32 16 8 4 2 1

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts +16: 198.16.40.0- 198.16.55.255

+16: 198.16.56.0- 198.16.71.255 etc

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000 /16

variable subnet mask: 11111111.11111111.11100000.000000 /19

now we have added 3 1s in the 3rd field to reserve 13 0s

subnet mask: 255.255.32.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1 1  1  1 1 1 1 1

128 64 32  16 8 4 2 1

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

Ist valid Ip add: 198.16.40.1 - 198.16.71.255(last valid IP address)

Company C starts +16: 198.16.72.0- 198.16.103.255

+16: 198.16.104.0- 198.16.136.255 etc

5 0

3 years ago

Whats my screen name?

adoni [48]

It could be any thing of your chose

8 0

3 years ago

Read 2 more answers