What is token passing?

modify hw 02 c.c to use envp instead of environ. be sure that you understand how the code works. provide liberal comments to exp

rjkz [21]

With the aid of pointer-based arithmetic operations and the usage of pointers in comparison operations, address arithmetic is a technique for determining the address of an object. Pointer arithmetic is another name for address arithmetic.

The pointers can be used for mathematical operations like addition, subtraction, etc. The outcome of an arithmetic operation on the pointer, however, will likewise be a pointer if the other operand is of type integer because we know that the pointer includes the address. These operations are addition and subtraction. In C++, a pointer's value can be increased or decreased. It signifies that we can change the pointer's value by adding or removing integer values. A pointer arithmetic can be subtracted (or added) from another in a manner similar to this.

Learn more about arithmetic here-

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4 0

1 year ago

class secretType { public: static int count; static int z; secretType(); secretType(int a); void print(); static void incrementY

DanielleElmas [232]

Answer:

The answer to this question can be given as:

In this class definition, there are two constructors.

Explanation:

In the class definition two constructors are different in type but first we explain constructor that can be as:

Constructor: constructor are special member functions whose task is to initialized as an object of its class.

Rule for defining constructor:

A constructor doesn’t have a return type not even void.

The name of the constructor must be the same as the class name.

A constructor is called automatically when a new instance of an object is created.

Type of constructor:

default constructor.
parameterized constructor.
copy constructor.

In the question there are two type of constructor is used that can be given as:

default constructor:

The constructor without any parameters is called a default constructor. This type of constructor is called automatically when a new instance of an object is created.

parametrized constructor:

In the parameterized constructor we use at least one parameter in the constructor that is called the parameterized constructor. In the parameterized constructor we can initialize each instance of the class with several values.

Example :

class AB //define class

{

String name; //define variable

AB() //default constructor

{

System.out.print("hello..."); //message.

}

AB(String name) //parametrized constructor.

{

this.name = name; //holding value in name variable.

}

class Main //define class

{

public static void main (String[] args) //main method

{

AB OB1 =new AB(); //creating class object.

AB ob2=new AB("XYZ");

System.out.print(ob2.name); //print value

}

Output:

hello... XYZ

5 0

4 years ago

There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou

Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:- n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex u ∈ V

$\rightarrow$ for each vertex v ∈ V

$\rightarrow\rightarrow$ if( d(pu, pj) > k )

$\rightarrow\rightarrow\rightarrow$ add vertex u to Adj[v] // Adj[v] represents adjacency list of v

$\rightarrow\rightarrow\rightarrow$ add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3. bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

$\rightarrow$ visited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph

6. for each vertex u ∈ V

$\rightarrow$ if ( visited[u] == false)

$\rightarrow\rightarrow$ color[u] = 0;

$\rightarrow\rightarrow$ isbipartite = BFS(G,u,color,visited) // if the vertices reachable from u form a bipartite graph, it returns true

$\rightarrow\rightarrow$ if (isbipartite == false)

$\rightarrow\rightarrow\rightarrow$ print " No solution exists "

$\rightarrow\rightarrow\rightarrow$ exit(0)

7. for each vertex u ∈V

$\rightarrow$ if (color[u] == 0 )

$\rightarrow\rightarrow$ print " Student u is assigned Bus 1"

$\rightarrow$ else

$\rightarrow\rightarrow$ print " Student v is assigned Bus 2"

BFS(G,s,color,visited)

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

$\rightarrow$ u = Q.pop()

$\rightarrow$ for each vertex v ∈ Adj[u]

$\rightarrow\rightarrow$ if (visited[v] == false)

$\rightarrow\rightarrow\rightarrow$ color[v] = (color[u] + 1) % 2

$\rightarrow\rightarrow\rightarrow$ visited[v] = true

$\rightarrow\rightarrow\rightarrow$ Q.push(v)

$\rightarrow\rightarrow$ else

$\rightarrow\rightarrow\rightarrow$ if (color[u] == color[v])

$\rightarrow\rightarrow\rightarrow\rightarrow$ return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0

3 years ago

2) The CPU is made from a silicon<br>a) Steel<br>b) Chip<br>c) pendrive

almond37 [142]

Answer:

Address buses are made up of a collection of wires connecting the CPU with main memory that is used to identify particular locations (addresses) in main memory. The width of the address bus (that is, the number of wires) determines how many unique memory locations can be addressed.

Explanation:

6 0

3 years ago

True or False

NISA [10]

Answer: The answer would be True

3 0

3 years ago