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skad [1K]
3 years ago
14

Examine the following declarations and definitions for the array-based implementations for the Stack and Queue ADTs. Assume that

exception class PushOnFullStack and class PopOnEmptyStack have been defined and are available. Read the following code segment and fill in blank #6.
class StackType
{

public:

StackType();

void Push(StackItemType item);

void Pop();

private:

int top;

ItemType items[MAX_STACK];

};

void StackType::StackType()

{

top = -1;

}

void StackType::Push(ItemType item)

{

__________________ // 1

___________________; // 2

__________________; // 3

___________________; // 4

}

class QueType

{

public:

// prototypes of QueType operations

// go here

private:

int front;

int rear;

ItemType items[MAX_QUEUE];

}

void QueType::QueType()

{

front = MAX_QUEUE - 1;

rear = MAX_QUEUE - 1;

}

Boolean QueType::IsEmpty()

{

return (rear == front);

}

void QueType::Enqueue(ItemType item)

{

____________________; // 5

____________________; // 6

}

[1] rear = (rear +1) % MAX_QUEUE
[2] items[rear] = item
[3] rear = (rear % MAX_QUEUE) + 1
[4] items[front] = item
Computers and Technology
1 answer:
otez555 [7]3 years ago
8 0

Answer:

The codes for the respective blanks are given below with appropriate comments for better understanding

Explanation:

<u>FOR 1 TO 4</u>

Void StackType::Push(ItemType item)

{

if(top == MAX_STACK - 1) // means stack is full so we need to throw the PushOnFullStack exception

throw PushOnFullStack ; // You can use this class and appropriate method to deal with exception like printing that stack is full so can not push the current item

top ++ ;// increment the top to accumulate the next item

items[top] = item; // put the item into the place identified

}

<u>FOR 5 AND 6</u>

Void Quetype::enqueue(itemType item)

{

if (rear==MAX_QUEUE && front==0) // queue is full so can not enqueue

//Handle the queue full exception here, may be print this

}else

{

items[rear] = item;

rear ++;

}

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Answer:

package Lab6Problema;

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{

System.out.println();

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System.out.println("******");

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public static void W()//prints W

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System.out.println("***** ");

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public static void D()//prints D

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System.out.println();

System.out.println("**** ");

System.out.println("* * ");

System.out.println("* *");

System.out.println("** * ");

System.out.println("**** ");

}

public static void blank() //prints blank space

{

System.out.println();

System.out.println();

System.out.println();

}

public static void main(String[] args) {

//just enter "Hello World" as input as specified

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for(int i=0;i<numOfCharacters;i++) // loop through each chracter in msg

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// gets character at position i, and switch accordingly

switch(myMessage.charAt(i))

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case 'H':

case 'h':H();break;

case 'E':

case 'e':E();break;

case 'L':

case 'l':L();break;

case 'O':

case 'o':O();break;

case 'W':

case 'w':W();break;

case 'R':

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case 'D':

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}

}

}

Explanation:

4 0
3 years ago
Which of the following is a programming language that permits Web site designers to run applications on the user's computer?
muminat

Answer:

c) Java.

Explanation:

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Answer:

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Explanation:

Solution

Penetration testing has several methodologies which include :OSSTMM and PTES  

The comparison between OSSTMM and PTES is stated as follows:

OSSTMM:                                                

Security assessment methodology

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PTES :

Technology oriented

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Now,

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Now,

The OSSTMM is used to obtain security metrics and performing penetration testing .The OSSTMM provides transparency to those who have inadequate security policies and configurations.

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