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2 years ago

A variable must have its type declared but it is not required to be initialized prior to first use.

Computers and Technology

1 answer:

AnnZ [28]2 years ago

3 0

Answer:

a) TRUE.

Explanation:

The statement is true.We can declare a variable but not initialize it there we can initialize it afterwards when we want to use.On declaring the variable type it tells the compiler to reserve the memory according to the size required for the data type.When we initialize the memory is allocated to that variable.

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State four features of information

Xelga [282]

Answer:

This is your answer

8 0

3 years ago

Select all the correct answers. Which two statements are true about an OS? translates the user's instructions into binary to get

lubasha [3.4K]

Answer:

all

Explanation:

4 0

3 years ago

What will be displayed after code corresponding to the following pseudocode is run? Main Set OldPrice = 100 Set SalePrice = 70 C

Tatiana [17]

Answer:

A jacket that originally costs $ 100 is on sale today for $ 80

Explanation:

Main : From here the execution of the program begins

Set OldPrice = 100 -> This line assigns 100 to the OldPrice variable

Set SalePrice = 70 -> This line assigns 70to the SalePrice variable

Call BigSale(OldPrice, SalePrice) -> This line calls BigSale method by passing OldPrice and SalePrice to that method

Write "A jacket that originally costs $ ", OldPrice -> This line prints/displays the line: "A jacket that originally costs $ " with the resultant value in OldPrice variable that is 100

Write "is on sale today for $ ", SalePrice -> This line prints/displays the line: "is on sale today for $ " with the resultant value in SalePrice variable that is 80

End Program -> the main program ends

Subprogram BigSale(Cost, Sale As Ref) -> this is a definition of BigSale method which has two parameters i.e. Cost and Sale. Note that the Sale is declared as reference type

Set Sale = Cost * .80 -> This line multiplies the value of Cost with 0.80 and assigns the result to Sale variable

Set Cost = Cost + 20 -> This line adds 20 to the value of Cost and assigns the result to Cost variable

End Subprogram -> the method ends

This is the example of call by reference. So when the method BigSale is called in Main by reference by passing argument SalePrice to it, then this call copies the reference of SalePrice argument into formal parameter Sale. Inside BigSale method the reference &Sale is used to access actual argument i.e. SalePrice which is used in BigSale(OldPrice, SalePrice) call. So any changes made to value of Sale will affect the value of SalePrice

So when the method BigSale is called two arguments are passed to it OldPrice argument and SalePrice is passed by reference.

The value of OldPrice is 100 and SalePrice is 70

Now when method BigSale is called, the reference &Sale is used to access actual argument SalePrice = 70

In the body of this method there are two statements:

Sale = Cost * .80;

Cost = Cost + 20;

So when these statement execute:

Sale = 100 * 0.80 = 80

Cost = 100 + 20 = 120

Any changes made to value of Sale will affect the value of SalePrice as it is passed by reference. So when the Write "A jacket that originally costs $ " + OldPrice Write "is on sale today for $ " + SalePrice statement executes, the value of OldPrice remains 100 same as it does not affect this passed argument, but SalePrice was passed by reference so the changes made to &Sale by statement in method BigSale i.e. Sale = Cost * .80; has changed the value of SalePrice from 70 to 80 because Sale = 100 * 0.80 = 80. So the output produced is:

A jacket that originally costs $ 100 is on sale today for $ 80

7 0

3 years ago

true or false? in a known-plaintext attack (kpa), the cryptanalyst hs access only to a segment of encrpted data and has no choic

bulgar [2K]

Yes , it’s true. In a known-plaintext attack (kpa), the cryptanalyst can only view a small portion of encrypted data, and he or she has no control over what that data might be.

The attacker also has access to one or more pairs of plaintext/ciphertext in a Known Plaintext Attack (KPA). Specifically, consider the scenario where key and plaintext were used to derive the ciphertext (either of which the attacker is trying to find). The attacker is also aware of what are the locations of the output from key encrypting. That is, the assailant is aware of a pair. They might be familiar with further pairings (obtained with the same key).

A straightforward illustration would be if the unencrypted messages had a set expiration date after which they would become publicly available. such as the location of a planned public event. The coordinates are encrypted and kept secret prior to the event. But when the incident occurs, the attacker has discovered the value of the coordinates /plaintext while the coordinates were decrypted (without knowing the key).

In general, a cipher is easier to break the more plaintext/ciphertext pairs that are known.

To learn more about Plaintext Attack click here:

brainly.com/question/28445346

#SPJ4

6 0

1 year ago

2. A host computer with an IP address of 172.16.29.155 using a subnet mask of 255.255.255.0 and a gateway of 172.16.29.1 pings a

ASHA 777 [7]

Answer:

From the host IP address and subnet mask, you can tell that both computers are on the same network. Host 172.16.29.155 will need to send an ARP request broadcast for 172.16.29.83 to get the MAC address and update its ARP table if the host 172.16.29.83 has not already done so.

Explanation:

<h2>ADDRESS RESOLUTION PROTOCOL (ARP)</h2>

Address Resolution Protocol (ARP) is a procedure for mapping a logical address known as the IP address to a physical address known as MAC (Media Access Control) address in a local area network. The job of the ARP is essentially to translate 32-bit addresses to 48-bit addresses and vice-versa when using IP Version 4 (IPv4) to communicate on the network.

Most of the computer programs/applications use logical address (IP address) to send/receive messages, however, the actual communication happens over the physical address (MAC address) this happens in the layer 2 of the OSI model. An ARP request is a broadcast, and an ARP response is a Unicast.

Scenarios when ARP is used:

Scenario-1: The sender is a host and wants to send a packet to another host on the same network.

· Use ARP to find another host’s physical address

Scenario -2: The sender is a host and wants to send a packet to another host on another network.

· Sender looks at its routing table.

· Find the IP address of the next-hop (router) for this destination.

· Use ARP to find the router’s physical address

Scenario -3: the sender is a router and received a datagram destined for a host on another network.

· Router checks its routing table.

· Find the IP address of the next router.

· Use ARP to find the next router’s physical address.

Scenario -4: The sender is a router that has received a datagram destined for a host in the same network.

· Use ARP to find this host’s physical address.

The situation in the question above falls under scenario-1. From the host IP address and subnet mask, you can tell that both computers are on the same network. Host 172.16.29.155 will need to send an ARP request broadcast for 172.16.29.83 so to get the MAC address and update its ARP table if the host 172.16.29.83 has not already done so.

7 0

3 years ago