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trasher [3.6K]
2 years ago
9

Which of the following is data temporarily stored on?

Computers and Technology
2 answers:
Bogdan [553]2 years ago
5 0

Answer:

The answer is memory chip

Explanation:

i took the test

olga nikolaevna [1]2 years ago
4 0

Answer:

Memory chip or hard drive

Explanation:

You might be interested in
Diagnosing is solving the problem, and trouble shooting is figuring out what the problem is.
exis [7]
False its the opposite
3 0
3 years ago
Characters used to control the position of the output on the screen or paper, to cause some action to occur, such as ringing a b
Sedbober [7]

Answer:

There are four options for this question, and the letter C is correct:

a) glyphs.

b) symbols.

c) control characters.

d) command characters.

Explanation:

Control characters is a code with a number a several characters, Procedural signs in Morse code are a form of control character, a form of control was introduced in 1870 called Baudot code: NUL and DEL, and was introduced in 1901 Murray code added the carriage return (CR) and line feed (LF).

For example:

(null, NUL, \0, ^@) used by many programming language like "C".

5 0
3 years ago
Modify the NumberedList class we implementd during the lecture by adding a member function: void NumberedList::insertPosition(in
satela [25.4K]

Answer:

Check the explanation

Explanation:

#include<iostream>

using namespace std;

class LinkedList

{

private:

   // Declare a structure for the list.

   struct ListNode

   {

   int value;        // The value in this node.

   struct ListNode *next;// To point to the next node.

   };

   ListNode *head;        // List head pointer.

public:

   // Constructor.

   LinkedList()

       { head = NULL; }

   

   // Destructor

   ~LinkedList();

   

   // Linked list operations.

   void appendNode( int );

   void insertNode( int );

   void insertNodeAt(int,int);

    void deleteNode( int );

   void Reverse();

   void deleteAt(int);

   int Search(int);

   void display() const;

};

// appendNode appends a node containing the      

// value passed into num, to the end of the list.  

void LinkedList::appendNode( int num )

{

  ListNode *newNode; // To point to a new node.

  ListNode *nodePtr; // To move through the list.

  // Allocate a new node and store num there.

  newNode = new ListNode;

  newNode->value = num;

  newNode->next = NULL;

  // If there are no nodes in the list.

  // make newNode the first node.

  if ( !head )

     head = newNode;

  else // Otherwise, insert newNode at end.

  {

     // Initialize nodePtr to head of list.

     nodePtr = head;

     // Find the last node in the list.

     while ( nodePtr->next )

        nodePtr = nodePtr->next;

     // Insert newNode as the last node.

     nodePtr->next = newNode;

  }    //    end else-if

  display();  

}    //    end function appendNode

// displayList shows the value stored in each              

// node of the linked list pointed to by head.      

                       

void LinkedList::display() const

{

   ListNode *nodePtr; // To move through the list

   if ( !head )

   {

       cout << "\n\tThe list is empty.";

       return;

   }

   // Position nodePtr at the head of the list.

   nodePtr = head;

   cout << "\n\n\tThe elements in the list are:\n\t";

   // While nodePtr points to a node, traverse the list.

 

    while (nodePtr)

   {

       // Display the value in this node.

       cout << nodePtr->value << " -> ";

       // Move to the next node.

       nodePtr = nodePtr->next;

   }    //    end while.

   cout << "Null";

}    //    end function displayList.

// Reverse function re-arranges node in the list.

void LinkedList::Reverse()

{

   ListNode *nodePtr;

   ListNode *next;

   ListNode *result=NULL;

   if ( !head )

   {

       cout << "\n\tThe list is empty.";

       return;

   }

   // Position nodePtr at the head of the list.

   nodePtr = head;

   while (nodePtr!=NULL)

   {

       next=nodePtr->next;

       nodePtr->next=result;

       result=nodePtr;

       nodePtr=next;

   }

   head=result;

display();

}

// The insertNode function inserts a node with num copied to its value member.                

void LinkedList::insertNode( int num )

{

   ListNode *newNode;             // A new node.

   ListNode *nodePtr;             // To traverse the list.

   ListNode *previousNode = NULL; // The previous node.

   // Allocate a new node and store num there.

   newNode = new ListNode;

   newNode->value = num;

   newNode->next = NULL;

 

   // If there are no nodes in the list make newNode the first node.

   if ( !head )

       head = newNode;

   else // Otherwise, insert newNode.

   {

       // Position nodePtr at the head of list.

       nodePtr = head;

       //    Initialize previousNode to NULL.

       previousNode = NULL;

       //    Skip all nodes whose value is less than num.

       while ( nodePtr != NULL && nodePtr->value < num )

       {

           previousNode = nodePtr;

           nodePtr = nodePtr->next;

       }

       //If the new node is to be the 1st in the list,

       //    insert it before all other nodes.

       if ( previousNode == NULL )

       {

           head = newNode;

           newNode->next = nodePtr;

       }

       else // Otherwise insert after the previous node.

       {

           previousNode->next = newNode;

           newNode->next = nodePtr;

       }

   }    //    end else-if

     

display();  

}    //    end function insertNode.

// The insertNode function inserts a node at pos  

//with num copied to its value member.          

void LinkedList::insertNodeAt( int num ,int pos)

{

   ListNode *newNode;             // A new node.

   ListNode *nodePtr;             // To traverse the list.

   ListNode *previousNode = NULL; // The previous node.

   int i=0;

   // Allocate a new node and store num there.

   newNode = new ListNode;

   newNode->value = num;

   newNode->next = NULL;

   // Position nodePtr at the head of list.

       nodePtr = head;

   if(pos==0)//to inserted at first.

   {  

       newNode->next=head;

       head=newNode;

   }

   else

   {

   while(nodePtr != NULL && i<pos) //loop to reach position.

       {  

           previousNode=nodePtr;

           nodePtr=nodePtr->next;

           i++;

       }

       if(nodePtr==NULL)//position not found.

           cout<<"Invalid Position :"<<endl;

       else//inserts node.

       {

           newNode->next=nodePtr;

           previousNode->next=newNode;

       }

   }

   display();

}

//    The deleteNode function searches for a node with num as its value.  

//The node, if found, is deleted from the list and from memory.

void LinkedList::deleteNode( int num )

{

   ListNode *nodePtr;       // To traverse the list.

   ListNode *previousNode;//To point to the previous node.

   // If the list is empty, do nothing.

   if ( !head )

   {

       cout << "\n\tFailed to delete as list is empty.";  

       return;

   }

   // Determine if the first node is the one.

   if ( head->value == num )

   {

       nodePtr = head->next;

       delete head;

       head = nodePtr;

   }

   else

   {

       // Initialize nodePtr to head of list.

       nodePtr = head;

       // Skip all nodes whose value member is not equal to num.

       while (nodePtr != NULL && nodePtr->value != num)

       {

           previousNode = nodePtr;

           nodePtr = nodePtr->next;

       }

3 0
3 years ago
New and just need help with C coding. I've tried if statements and it outputs the wrong number.
dimaraw [331]

Answer:

#include <stdio.h>

int main(void) {

 

 int num1;

 int num2;

 int num3;

 printf("Enter three integers: ");

 scanf("%d", &num1);

 scanf("%d", &num2);

 scanf("%d", &num3);

 if (num1 == 0 || num2 == 0 || num3 == 0)

 {

   printf("please input a number greater than zero :)\n");

 }

 if (num1 <= num2 && num1 <= num3)

 {

   printf("%i is the smallest number!\n", num1);

 }

 else if (num2 <= num1 && num2 <= num3)

 {

   printf("%i is the smallest number!\n", num2);

 }

 else

 {

   printf("%i is the smallest number!\n", num3);

 }

 return 0;

}

Explanation:

Alright so let's start with the requirements of the question:

  1. must take 3 integers from user input
  2. determine which of these 3 numbers are the smallest
  3. spit out the number to out

So we needed to create 3 variables to hold each integer that was going to be passed into our script.

By using scanf("%i", &variableName) we were able to take in user input and  store it inside of num1, num2, and num3.

Since you mentioned you were new to the C programming language, I threw in the first if statement as an example of how they can be used, use it as a guide for future reference, sometimes it's better to understand your code visually.

Basically what this if statement does is, it checks to see if any of the integers that came in from user input was the number zero, it told the user it does not accept that number, please input a number greater than zero.

if (num1 == 0 || num2 == 0 || num3 == 0)

 {

   printf("please input a number greater than zero :)\n");

 }

I used this methodology and implemented the heart of the question,

whichever number is smaller, print it out on the shell (output).

if (num1 <= num2 && num1 <= num3)

^^ here we're checking if the first variable we created is smaller than the second variable and the third ^^

 {

   printf("%i is the smallest number!\n", num1);

   ^^ if it is smaller, then print integer and then print a new line so the next line looks neat ^^

   ( incase if your wondering what "\n" is, its a special character that allows you so print a new line on the terminal, kind of like hitting the return or enter key )

 }

else if (num2 <= num1 && num2 <= num3)

^^ else if is used when your checking for more than one thing, and so for the second variable we checked to see if it was smaller than the first and third variable we created ^^

 {

   printf("%i is the smallest number!\n", num2); < -- and we print if it's smaller

 }

Last but not least:

else

^^ if it isn't num1 or num2, then it must be num3 ^^

 {

   printf("%i is the smallest number!\n", num3);

  we checked the first two options, if its neither of those then we have only one variable left, and thats num3.

 }

I hope that helps !!

Good luck on your coding journey :) ‍

7 0
2 years ago
Read 2 more answers
Secondary hard drive whats its purpose
loris [4]
As a back up in cases like the first hard drive gets fried
4 0
3 years ago
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