Beyond aesthetics- what functional differences are there for running G.docx
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Beyond aesthetics, what functional differences are there for running GNS3 as root versus running GNS3 as a non-root user? Solution Differences between root and non root installations : During a root installation, subdirectories and files for the DB2® database product are created in a directory of the root user\'s choosing. Unlike root users, non-root users cannot choose where DB2 database products are installed. Non-root installations are always placed in the $HOME/sqllibdirectory, where $HOME represents the non-root user\'s home directory. The layout of the subdirectories within the sqllib directory of a non-root is similar to that of a root installation. For root installations, multiple instances can be created. Instance ownership is associated with the user ID under which the instance was created. Non-root installations can have only one DB2 instance. The non-root installation directory contains all of the DB2 database product files and instance files with no soft links. 1. In Root installations User can select installation directory. But in non root installations user can not select installation directory as DB2 database products are installed under the user\'s home directory. 2. In Root installations Multiple Number of DB2 instances are allowed, where as only one DB2 instances is allowed in non root installating proces. 3. Only Program files are deployed during installation in root installing, Instances must be created after installation. but in non root Program files and instance files both are deployed.The DB2 database product is ready for use immediately after installation. 4. No Upgrade version and instances are available in root installation process, No need to uninstall the old version before installing the new version in non root. Install new version, and upgrade the instance, together. .
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Beyond aesthetics- what functional differences are there for running G.docx
Beyond aesthetics, what functional differences are there for running GNS3 as root versus running GNS3 as a non-root user? FUNCTIONAL DIFFERENCES FOR GNS3 NOT DB2.
Solution
During a root installation, subdirectories and files for the DB2 database product are created in a directory of the root users choosing
unlike root users, non root users cannot choose where DB2 database products are installed.Nonroot installations are always placed in the $HOME/sqllib directory where $HOME represents the non root users home directory.The layout of the subdirectories within the sqllib directory of a non root is similar to that of a root installation
For root installations, multiple instances can be created.Instance ownership is associated with the user id under which the instance was created
Non root installations can have only one DB2 instance.The non root installation directory contains all of the DB2 database product files and instance files with no soft links.
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directory
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Beyond aesthetics- what functional differences are there for running G.docx
Beyond aesthetics, what functional differences are there for running GNS3 as root versus running GNS3 as a non-root user? FUNCTIONAL DIFFERENCES FOR GNS3 NOT DB2.
Solution
During a root installation, subdirectories and files for the DB2 database product are created in a directory of the root users choosing
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c#
Write an application that displays the following patterns separately. Use for loop to generate the patterns. All asterisks (*) should be displayed by a single statement of the form Console.Write(“*â€);
(a)
*
**
***
****
*****
******
*******
********
*********
**********
(b)
**********
*********
********
*******
******
*****
****
***
**
*
(c)
**********
*********
********
*******
******
*****
****
***
**
*
(d)
*
**
***
****
*****
******
*******
********
*********
**********
Part 2: Modify your program to display all four patterns side by side.
Solution
Answer
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace SampleFloydConcept
{
class Floyd
{
static void Main(string[] args)
{
int numOfRows = 5;
int i, j, k ;
Console.WriteLine(\"(a)\");
for (i = 1; i <= numOfRows; i++)
{
for (j = 1; j <= numOfRows-i; j++)
{
// Console.Write(\"\");
}
for (k = 1; k <= i; k++)
{
Console.Write(\"*\");
}
Console.WriteLine(\"\");
}
Console.ReadLine();
Console.WriteLine(\"(b)\");
for (i = 1; i <= numOfRows; i++)
{
for (j = 1; j <= numOfRows-i; j++)
{
// Console.Write(\" \");
}
for (k = 1; k <= j; k++)
{
Console.Write(\"*\");
}
Console.WriteLine(\"\");
}
Console.ReadLine();
Console.WriteLine(\"(d)\");
for (i = 1; i <= numOfRows; i++)
{
for (j = 1; j <= numOfRows-i; j++)
{
Console.Write(\" \");
}
for (k = 1; k <= i; k++)
{
Console.Write(\"*\");
}
Console.WriteLine(\"\");
}
Console.ReadLine();
}
}
}
.
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Write an application that displays the following patterns separately. Use for loop to generate the patterns. All asterisks (*) should be displayed by a single statement of the form Console.Write(“*â€);
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Part 2: Modify your program to display all four patterns side by side.
Solution
Answer
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace SampleFloydConcept
{
class Floyd
{
static void Main(string[] args)
{
int numOfRows = 5;
int i, j, k ;
Console.WriteLine(\"(a)\");
for (i = 1; i <= numOfRows; i++)
{
for (j = 1; j <= numOfRows-i; j++)
{
// Console.Write(\"\");
}
for (k = 1; k <= i; k++)
{
Console.Write(\"*\");
}
Console.WriteLine(\"\");
}
Console.ReadLine();
Console.WriteLine(\"(b)\");
for (i = 1; i <= numOfRows; i++)
{
for (j = 1; j <= numOfRows-i; j++)
{
// Console.Write(\" \");
}
for (k = 1; k <= j; k++)
{
Console.Write(\"*\");
}
Console.WriteLine(\"\");
}
Console.ReadLine();
Console.WriteLine(\"(d)\");
for (i = 1; i <= numOfRows; i++)
{
for (j = 1; j <= numOfRows-i; j++)
{
Console.Write(\" \");
}
for (k = 1; k <= i; k++)
{
Console.Write(\"*\");
}
Console.WriteLine(\"\");
}
Console.ReadLine();
}
}
}
.
Business Law 1- Firm X and Firm Y are in dispute over a contract- They.docx
Business Law
1. Firm X and Firm Y are in dispute over a contract. They decide to use a form of alternative dispute resolution (ADR) to settle the dispute in which both parties meet to discuss the issue. Firm X brings their lawyer to this meeting but Firm Y does not. Which form of ADR is most likely being used here?
1. Negotiation
2. Mediation
3. Arbitration
4. Facilitation
2. Which among the following does a binding arbitration clause typically state?
1. Why arbitration is being chosen over other methods of ADR
2. Which specific type of dispute is to be resolved by arbitration
3. How the arbitrator will be selected
4. What to do if a party is dissatisfied with the solution
3. David Myer seeks to resolve his dispute with his manager Ray Thorn. They both zero in on Rachel Moran to act as a neutral third party. Myer and Thorn take turns to explain their position to Ms. Moran who enumerates the strengths and weaknesses of the case. Using this assessment, Myer and Thorn reach a settlement. Which form of ADR is being used here?
1. Med-Arb
2. Peer review
3. Private trial
4. Early neutral case evaluation
Solution
1. Firm X and Firm Y are in dispute over a contract. They decide to use a form of alternative dispute resolution (ADR) to settle the dispute in which both parties meet to discuss the issue. Firm X brings their lawyer to this meeting but Firm Y does not. Which form of ADR is most likely being used here?
1. Negotiation
2. Which among the following does a binding arbitration clause typically state?
3. How the arbitrator will be selected
3. David Myer seeks to resolve his dispute with his manager Ray Thorn. They both zero in on Rachel Moran to act as a neutral third party. Myer and Thorn take turns to explain their position to Ms. Moran who enumerates the strengths and weaknesses of the case. Using this assessment, Myer and Thorn reach a settlement. Which form of ADR is being used here?
3. Private trial
.
Brief explain how a DBMS handles concurrent execution of transactionsS.docx
Brief explain how a DBMS handles concurrent execution of transactions
Solution
CONCURENT EXECUTION :
Schedule is a collection of one or more transactions and are executed according to their arrangement in the schedule.Schedule is catogorized into 2 divisions.They are : a)Serial and b) Concurrent
**//First let me introduce you Serial and later explanation on Concurrent Execution//**
a) Serial : An execution of transactions occurs one after the other(one-by-one)
b) Concurrent Execution : It is nothing but an execution of transactions parallelly by sharing the CPU time among them.The database has the chances of containing inconsistence value if the concurrent execution is not considered properly.
Let me explain you with an example,
Consider two transactions T1 and T2 which follows the instructions as below:
T1 :
Read S;
S = S*100;
Write S;
Read F;
F=F/2;
Write F;
T2 :
Read S;
Temp = S * 12;
Read B;
B = B +800;
Write B;
If T2 is a new transaction and suppose B should be deposited with the Rs.800 which is to be deduced from S in Transaction T1.
If we want a serial schedule for Transactions T1 and T2 then either T1 executes before the T2 gets executed or else T1 will execute before T2 is being executed..So now by using Context switching in concurrent Execution Transaction T1 gets executed partially and later T2 also partially.
The following example can support these lines
T1 T2
Read S;
S = S *100;
Write A;
Read S;
Temp = S *12;
Read B;
B = B + Temp;
Write B;
Read F;
F = F /2;
Write F;
As we used Context Switching in this type of Schedule, the 1st one after performing the 3rd line of Transaction T1, and after performing the last statement of T2. T1 first multiplies the amount with 12 with S and writes the new value into S.Later T2 reads the value of S, calculates the value of Temp to be Rs.250(say) and adds the value to B. The leftout part of T1 is executed and Rs.250 is added to B
So it is clear that the correct Context Switching is very important to maintain ACID properties like Atomicity,Consistency,Isolation and Durability.
.
Briefly explain the three types of IPv6 addresses-SolutionThree types.docx
Briefly explain the three types of IPv6 addresses.
Solution
Three types of IPv6 (Internet Protocol version 6 )addresses.
An IPv6 label is used to identify a network interface of a computer or other network node participating in an IPv6 computer network. IPv6 is the successor to the first addressing infrastructure of the internet , IPv4(internet protocol version 4). IPv4 IP address has 32-bit value where as IPv6 addresses have a size of 128-bit vale. That is why IPv6 has enlarged address space compared to IPv4.An IP address serves the purpose of uniquely identifying an individual n/w interface of a host, which is locating on the network, and it permits the routing of IP packets between hosts. IP addresses are present in the field of packet header. For routing IP addresses indicates source and destination of the packet.
The following are the three types of IP addresses which are supported by IPv6
IPv6 addresses are classified by the primary addressing and routing methodologies common in networking. They are ………
This is an identifier for a single interface. A Packet sent to a unicast address is delivered to the interface identified by that address. It can be link-local scope, site-local scope,or global scope.
Ths is an identifier for a group of interfaces generally belonging to different nodes. A packet sent to multicast address is delivered to all interfaces identified by that address.
An identifier for a group of interfaces generally belonging to different nodes. A packet sent to an anycast address is delivered to the closest member of a group or nearest member of a group, according to the routing protocols measure of distance.
Anycast addresses are taken from the unicast address spaces and are not syntactically distinguishable from unicast addresses. Anycast is described as a cross between unicast and multicast. Like multicast, multiple nodes may be listening on an anycast address. Like unicast, a packet sent to an anycast address will be delivered to only one of those nodes. The exact node to which is delivered is based on the IP routing table in the n/w.
Unicast and anycast addresses are composed of two logical parts.
A 64-bit n/w prefix used for routing and a 64-bit interface identifier used to identify host’s n/w interface. The n/w prefix is the combination of routing prefix combined with subnet id. It contained in the most significant 64 bits of the address. The size of the routing prefix may vary. A larger prefix size means a smaller subnet id size. The 64-bit interface identifier is either automatically generated of assigned manually.
Multcast addresses are formed according to several specific formatting rules, depending on the application.
.
Brief timeline of the development of the theory and or technology and.docx
Brief timeline of the development of the theory and or technology and key players of the physics of Particle and Wave detection
Solution
Please refer the below list to understand timeline of fundamental physics discoveries.
1. In 250BE - Archimedes principle by Archimedes.
2. In 1514 - Heliocentrism by Nicholas Copernicus.
3. In 1589 - Galileo leaning tower of Pisa experiment by Galileo Galilei.
4. In 1621 - Snell law to define refraction by willebrord snellius.
5. In 1660 - Pascal law to define properties of pressure by Pascal
6. In 1660 - Hooks law to define the relationship between stress and strain by Robert Hooks.
7. In 1687 - law of motion to define the concept of force and inertia and most important part gravitational law by Issac Newton.
8. In 1785 - Inverse square law to define the force between the charges by Charles coulomb.
9. In 1801 - wave theory of light by Thomas young.
10. In 1806 - Kinetic energy of the particle moving with some velocity by Thomas young.
11. In 1814 - wave theory of light explaining interference by Fresnel dishouffer.
12. In 1827 - Electrical Resistance (properties of conductor) by Ohm
13. In 1831 and 1838 - EMI (Electromagnetic induction) explaining the concept of production of induced emf and electrical lines of force by Faraday.
14. In 1851 - second law of thermodynamic which give the idea of heat engine and refrigerator by Rudolf, Kelvin and Clausius.
15. In 1861 - Black Body radiation by Maxwell.
16. In 1863 - Entropy by Clausius.
17. In 1867 - theory of gases by James clerk Maxwell.
18. In 1884 - Boltzmann theory which explain radiation law.
19. In 1887 - electromagnetic wave by Hertz.
20. In 1895 - X rays by Rontgen
21. In 1897 - Electron (charge particle ) by JJ Thomson.
22. In 1905 - Relativity by Newton
23. In 1913 - Bohr atomic model to define atom by Bohr.
24. In 1927 - Big bang theory which explain the formation of universe by George\'s.
25. In 1927 - Uncertainty principles - by Heisenberg.
26. In 2015 - Gravitional detected.
.
Both Ford and Phillips Electric have faced major changes globally in t.docx
Both Ford and Phillips Electric have faced major changes globally in their markets and industry structures with the emergence of new competitors, new markets and new technologies. Historically both companies pursued a multi-domestic approach to managing their global operations but have had to adapt given new circumstances. Compare the two companies responses to these developments and why one seems to be more successful in making the transition to a transnational approach.
Solution
Increase in globalization and industrialization has created an intense competition at the marketplace. In order to sustain and grow in such environment companies require continuous adaptation of new strategies.
Companies utilizing a transnational approach possess well-coordinated business activities between their head office and local and international operational divisions. It helps them to utilize various benefits including centralization of business activities and increasing the local responsiveness.
Ford is one of the largest producers of automobiles across the globe. Its major production facilities are located in Europe, America, Australia and South Africa. Besides this it has established affiliates in more than 95 countries.
Initially the company was utilizing decentralized strategy . Its operational business units in different nations used to have different decision making units in different countries. Later on, the company restructured its business model to become centralized. Besides this, it has stated acquiring different companies. This strategy has helped the company to increase its market share and growth. It also increased the productivity and sales revenue of the company. Ford could successfully utilize this structure due to its flexible business operations and ability to handle the structural change .
Contrary to this, Philips could not utilize this approach as they lacked the flexibility to successfully operate in global environment . Major reason behind this failure was larger dependency of the company in its centralized operations .
.
Brand Foods Wholesale Corporation operates more than 590 membership wa.docx
Brand Foods Wholesale Corporation operates more than 590 membership warehouses and employs more than 164,000 people. Its annual report included the following items at August 31, 2011 (in millions of dollars) %) (Click the icon to view the data.) Requirement 1. Prepare a condensed balance sheet including amounts for the following a. Inventory b. Total stockholders\' equity c. Total liabilities Brand Foods Wholesale Corporation Balance Sheet August 31, 2011 (in millions) Assets Liabilities and Stockholders\' Equity $ 4,024 Accounts payable Cash and cash equivalents Accounts receivable Inventories Property, net of accumulated depreciation Other assets S 6,582 2,097 1,258 4.257 996 Accrued liabilities 6,640 Long-term debt 12,369 Other liabilities 2,751 Total liabilities $ 14,194
Solution
Condensed Balance Sheet:
a. Inventories: Total Assets - Cash and Cash Equivalents - Account Receivables - Property Net of Accumulated Dep. - Other Assets = Inventories
b. Total Stockholder\'s Equity: Retained Earning + Other Stockholder\'s Equity + Common Stock
c. Total Liabilities: Total Assets - Total Stockholder\'s Equity = Total Liabilities
Other datas are given in Question.
.
Betty retires from the BS Partnership when the basis of her one-third.docx
Betty retires from the BS Partnership when the basis of her one-third interest is $105,000, which includes her share of liabilities. At the time of her retirement, the partnership had the following assets:
Assets                         P/S Basis                    FMV
Cash                           $145,000                    $145,000
Receivables                  40,000                       40,000
Land 130,000 220,000
315,000Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 405,000
The partnership has $60,000 of liabilities when Betty retires. The partnership will pay Betty cash of $130,000 to retire her partnership interest.
What are the amount and character of the gain or loss for Betty?
Solution
Capital account of Betty:
Calculation of Revaluation profit:
Revaluation profit on Land = Fair Market value of land - Book value of Land = $220000 - $ 130000 = $ 90000
Share of Betty in Revaluation profit = Revaluation profit * her share = $ 90000 * 1/3 = $ 30000
Thus, we find that Betty incurrs a loss of $ 5000. This loss is of Business nature, i.e., revenue nature.
.
Below is accounting information for Cascade Company for 2013- What we.docx
Cascade Company had total assets of $699,000 at the end of 2013. Their revenue was $430,000 with expenses of $307,000. Their assets included $127,000 in cash, $254,000 in equipment, $42,000 in accounts receivable, and $76,000 in notes receivable.
Below is a depiction of a doubly-linked list implementation of the bag.docx
Below is a depiction of a doubly-linked list implementation of the bag ADT with two sentinels -one at the head and one at the tail. Strict DLink { TYPE value; strict DLink * next; struct DLink * prev;}; strict linked List { int size; struct DLink *frontSentinel; struct DLink *backSentinel;}; Write a function that at first reverses the list and then prints the values of the links in the list from front to back. Don\'t redirect to any other function. You must use assertions if necessary.
Solution
/* Program to reverse a doubly linked list */
#include <stdio.h>
#include <stdlib.h>
/* a node of the doubly linked list */
struct node
{
int data;
struct node *n;
struct node *pr;
};
/* Function to reverse a Doubly Linked List */
void reverse(struct node **head_r)
{
struct node *t = NULL;
struct node *cur = *head_r;
/* swap next and prev for all nodes of
doubly linked list */
while (cur != NULL)
{
t = cur->pr;
cur->pr = cur->n;
cur->n = t;
cur = cur->pr;
}
/* Before changing head, check for the cases like empty
list and list with only one node */
if(t != NULL )
*head_r = t->pr;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginging of the Doubly Linked List */
void push(struct node** head_r, int new_data)
{
/* allocate node */
struct node* n_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
n_node->data = n_data;
/* since we are adding at the begining,
prev is always NULL */
n_node->pr = NULL;
/* link the old list off the new node */
n_node->n = (*head_r);
/* change prev of head node to new node */
if((*head_r) != NULL)
(*head_r)->pr = n_node ;
/* move the head to point to the new node */
(*head_r) = n_node;
}
/* Function to print nodes in a given doubly linked list
This function is same as printList() of singly linked lsit */
void printList(struct node *node)
{
while(node!=NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
}
/* Drier program to test above functions*/
int main()
{
/* Start with the empty list */
struct node* head = NULL;
/* Let us create a sorted linked list to test the functions
Created linked list will be 10->8->4->2 */
push(&head, 2);
push(&head, 4);
push(&head, 8);
push(&head, 10);
printf(\"\ Original Linked list \");
printList(head);
/* Reverse doubly linked list */
reverse(&head);
printf(\"\ Reversed Linked list \");
printList(head);
getchar();
}
.
caiorie Potenhally useful information- 4-1843I K -C + 273-15 760 mmHg-.docx
caiorie Potenhally useful information. 4.1843I K -C + 273.15 760 mmHg= 1 atm 16 ounces-1 lb STP is exactly 1 atm and 0 1 Calorie 1000 calories 2.205 lbs. 1 kg 1. What is the correct name for P2Ss? a. phosphorus sulfide b. diphosphorus pentasulfide c. diphosphorus (V) sulfide d. diphosphorus tetrasulfide e. potassium sulfide What is the correct formula for tin (IV) sulfate? a. Sn(SO4h b. Sn(SO4) c. Sn (SO)s d. Sn (SO4)s e. SnSO 2. 3. Which is the correct formula for the polyatomic anion carbonate? ?.? b.Co, 2. ?. ??2 2. e. CO23 4. What is the formula of the binary compound formed from calcium and sulfur? a. CaS b. CaS2 c. CaSO4 d. Ca(SO4)2 e. These elements don\'t form a binary compound. 5. What is the oxidation state of iron in FeSO c. 0 d. +1 e, +2
Solution
Q1)
P2S5 (There are two phosphorus connected to five sulphur atoms)
Hence the name of the compound will be diphosphorus pentasulfide which is Option B
Q2)
SO4 has an oxidation state of -2 on the whole ion, since Sn is present in +4 oxidation state ,hence one Sn will be connected to two SO4(2-)
Therefore, final answer will be Sn(SO4)2 which is option a
Q3)
Correct answer is Option B
In carbonate ion, carbon is present in +4 oxidation state, which is in this compound, since oxygen is present in -2 oxidation state
CO3(2-) is correct choice
Q4)
Correct answer is Option A
Calcium generally exists in +2 and sulphur in -2 oxidation state, hence one calcium will be attached to one sulphur
Hence the formula will be CaS
Q5)
SO4 exists in -2 oxidation state
Hence Fe will be in + 2 state, since the overall compound is neutral in nature
Correct answer is Option E
.
C++ inheritance True or false- If A is the superclass and B is the sub.docx
C++ inheritance
True or false: If A is the superclass and B is the subclass, behaviour from B is available in A. Briefly explain your answer. (b) The University is a good example of a large and relatively complex system with many different types of entities. Students are either undergraduate or postgraduate, and pan-time or full-time. There is more than one type of staff member, including part-time and full-time. There are academic staff members, who conduct research and teach. There are research staff, who mostly conduct research and may do a little teaching. There are professional staff who provide administrative support and neither teach nor conduct research. Finally, there are the senior executives and management layers of the University, including the Vice-Chancellor and the CIO as examples. Arrange the following classes in an object hierarchy as a diagram to show where behaviour is shared and where it is extended: University, Person, Staff, Student, UnderGrad, PostGrad, Academic, Research, Professional, Executive (c) You have defined a class Animal and now wish to define a new class Panda that inherits from Animal and adds a single public function with this signature: void growl (); Write the C++ declaration for Panda that clearly shows the inheritance from Animal. Make sure that you explicitly declare the default constructor. (d) In the previous part, Animal has an integer variable weight defined as part of the class. What is the access modifier that must be used with weight if we want it available to Panda but not to other classes who haven\'t inherited Animal?
Solution
Answer :
A) False
When Class A is Super class and Class B is declred as subclass, then only Class B inherits features of super class A and also class B features can not be accessed in class A
E.g.
C) Class Animal is Super class and Class Panda is declred as subclass, which is using Single Inheritance
E.g.
D) When we delare variable weigth under access modifier protected in Class Animal then only Class Panda that inherits from Class Animal can access this variable.
E.g.
#include <iostream>
class Animal // Super class
{
public:
int legs;
protected:
int weigth;
Animal()
{
legs=4;
weigth=10;
}
};
class Panda: public Animal // Sub class
{
public:
void growl()
{
std::cout<<\"legs : \"<<legs<<\" weigth : \"<<weigth;
}
};
int main()
{
Panda obj;
obj.growl();
return 0;
}
.
C++ help! Write a function merge that merges two lists into one- alter.docx
C++ help! Write a function merge that merges two lists into one, alternating ele- ments from each list until the end of one of the lists has been reached, then append- ing the remaining elements of the other list. For example, merging the lists containing A B C and D E F G H should yield the list A D B E C F G H. However, do not create a third list; insert values of the second list into the first list. Your function should not fail when one (or both) of the lists is (are) empty. Write and test your merge function in a main function.
Please help. Thanks a ton!
Solution
hey heres the code , when you said \'lists\' I used STL, if you think I need to implement linked lists using struct then post comment.
#include <iostream>
#include <list>
using namespace std;
void merge(list<char>,list<char>);
// Simple example uses type int
int main()
{
list<char> L;
/*list 1 elements A B C */
L.push_front(\'A\');
L.push_back(\'B\');
L.push_back(\'C\');
list<char> K;
/*list 2 elements D E F G H */
K.push_back(\'D\');
K.push_back(\'E\');
K.push_back(\'F\');
K.push_back(\'G\');
K.push_back(\'H\');
cout << endl;
/*Calling merge function */
merge(L,K);
return 0;
}
void merge(list<char> A,list<char>B)
{
list<char>::iterator i;
list<char>::iterator k;
/*here it will go through loop to merge second list with first list*/
i=A.begin();
for(k=B.begin(); k!=B.end(); ++k)
{
if(i!=A.end() && k!=B.end())
{
if((++i)!=A.end())
A.insert(i,*k);Â Â /*insert function of list is used to insert element at position mentioned in first argument i (here we first incremented i and then we check if its not end of list then insert element)*/
else
A.push_back(*k);
}
else if(i==A.end() || k!=B.end())
{
A.push_back(*k);
}
else
break;
}
cout<<endl;
for(i=A.begin(); i != A.end(); ++i) cout << *i << \" \";
}
Output : A D B E C F G H
.
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- 1. Beyond aesthetics, what functional differences are there for running GNS3 as root versus running GNS3 as a non-root user? Solution Differences between root and non root installations : During a root installation, subdirectories and files for the DB2® database product are created in a directory of the root user's choosing. Unlike root users, non-root users cannot choose where DB2 database products are installed. Non- root installations are always placed in the $HOME/sqllibdirectory, where $HOME represents the non-root user's home directory. The layout of the subdirectories within the sqllib directory of a non-root is similar to that of a root installation. For root installations, multiple instances can be created. Instance ownership is associated with the user ID under which the instance was created. Non-root installations can have only one DB2 instance. The non-root installation directory contains all of the DB2 database product files and instance files with no soft links. 1. In Root installations User can select installation directory. But in non root installations user can not select installation directory as DB2 database products are installed under the user's home directory. 2. In Root installations Multiple Number of DB2 instances are allowed, where as only one DB2 instances is allowed in non root installating proces. 3. Only Program files are deployed during installation in root installing, Instances must be created after installation. but in non root Program files and instance files both are deployed.The DB2 database product is ready for use immediately after installation.
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