C++ PROGRAMThis program builds on the code below#include iostr.pdf

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C++ PROGRAM This program builds on the code below: #include #include #include \"shirtType.h\" using namespace std; void getInput(shirtType s[],ifstream& infile, int& numShirts); int main() { int numShirts; shirtType shirts[500]; ifstream infile; infile.open(\"h:\\\\shirtlab.txt\"); getInput(shirts,infile,numShirts); for(int i =0; i>id; infile.ignore(100,\'\ \'); getline(infile,c); infile>>size>>p>>q; while(infile) { s[index].setAll(id, c, size, p, q); index++; infile>>id; infile.ignore(100,\'\ \'); getline(infile,c); infile>>size>>p>>q; } numShirts=index; } See also the description of the class shirtType below: #include #include using namespace std; class shirtType { private: int shirtID; string color; string size; double price; int quantity; public: void setAll(int i,string c,string s, double p, int q); void print() const; void newQuan(); shirtType(); shirtType(int,string,string,double,int); }; shirtType::shirtType(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } shirtType::shirtType() { shirtID=0; color=\"\"; size=\"\"; price=0.0; quantity=0; } void shirtType::setAll(int i,string c,string s, double p, int q) { color = c; size = s; if(i>0) shirtID=i; else shirtID=0; if(p>0.0) price = p; else price=0.0; if (q>=0) quantity=q; else quantity = 0; } void shirtType::newQuan() { cout<<\"Enter the new quantity of shirts: \"<>quantity; } void shirtType::print() const { cout<<\"Shirt ID: \"< Solution #include #include struct node { int data; struct node *next; }*head; void append(int nkkm) { struct node *temp,*right; temp= (struct node *)malloc(sizeof(struct node)); temp->data=nkkm; right=(struct node *)head; while(right->next != NULL) right=right->next; right->next =temp; right=temp; right->next=NULL; } void add( int nkkm ) { struct node *temp; temp=(struct node *)malloc(sizeof(struct node)); temp->data=nkkm; if (head== NULL) { head=temp; head->next=NULL; } else { temp->next=head; head=temp; } } void addafter(int nkkm, int loc) { int i; struct node *temp,*left,*right; right=head; for(i=1;inext; } temp=(struct node *)malloc(sizeof(struct node)); temp->data=nkkm; left->next=temp; left=temp; left->next=right; return; } void addaShirt(int nkkm) { int c=0; struct node *temp; temp=head; if(temp==NULL) { add(nkkm); } else { while(temp!=NULL) { if(temp->datanext; } if(c==0) add(nkkm); else if(cdata==nkkm) { if(temp==head) { head=temp->next; free(temp); return 1; } else { prev->next=temp->next; free(temp); return 1; } } else { prev=temp; temp= temp->next; } } return 0; } void display(struct node *r) { r=head; if(r==NULL) { return; } while(r!=NULL) { printf(\"%d \",r->data); r=r->next; } printf(\"\ \"); } int count() { struct node *n; int c=0; n=head; while(n!=NULL) { n=n->next; c++; } return c; } int main() { int i,nkkm; struct node *n; head=NULL; while(1) { printf(\"1.Add a shirt\ \"); printf(\"2.Sell a shirt\ \"); printf(\"3.Res.

Education

{
private:
int shirtID;
string color;
string size;
double price;
int quantity;
public:
void setAll(int i,string c,string s, double p, int q);
void print() const;
void newQuan();
shirtType();
shirtType(int,string,string,double,int);
};
shirtType::shirtType(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
shirtType::shirtType()
{

shirtID=0;
color="";
size="";
price=0.0;
quantity=0;
}
void shirtType::setAll(int i,string c,string s, double p, int q)
{
color = c;
size = s;
if(i>0)
shirtID=i;
else
shirtID=0;
if(p>0.0)
price = p;
else
price=0.0;
if (q>=0)
quantity=q;
else
quantity = 0;
}
void shirtType::newQuan()
{
cout<<"Enter the new quantity of shirts: "<>quantity;
}
void shirtType::print() const
{
cout<<"Shirt ID: "<
Solution
#include
#include

struct node
{
int data;
struct node *next;
}*head;
void append(int nkkm)
{
struct node *temp,*right;
temp= (struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
right=(struct node *)head;
while(right->next != NULL)
right=right->next;
right->next =temp;
right=temp;
right->next=NULL;
}
void add( int nkkm )
{
struct node *temp;
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
if (head== NULL)
{
head=temp;
head->next=NULL;
}
else
{

temp->next=head;
head=temp;
}
}
void addafter(int nkkm, int loc)
{
int i;
struct node *temp,*left,*right;
right=head;
for(i=1;inext;
}
temp=(struct node *)malloc(sizeof(struct node));
temp->data=nkkm;
left->next=temp;
left=temp;
left->next=right;
return;
}
void addaShirt(int nkkm)
{
int c=0;
struct node *temp;
temp=head;
if(temp==NULL)
{
add(nkkm);
}
else
{
while(temp!=NULL)
{
if(temp->datanext;
}

if(c==0)
add(nkkm);
else if(cdata==nkkm)
{
if(temp==head)
{
head=temp->next;
free(temp);
return 1;
}
else
{
prev->next=temp->next;
free(temp);
return 1;
}
}
else
{
prev=temp;
temp= temp->next;
}
}
return 0;
}
void display(struct node *r)
{
r=head;
if(r==NULL)
{
return;
}
while(r!=NULL)
{

printf("%d ",r->data);
r=r->next;
}
printf(" ");
}
int count()
{
struct node *n;
int c=0;
n=head;
while(n!=NULL)
{
n=n->next;
c++;
}
return c;
}
int main()
{
int i,nkkm;
struct node *n;
head=NULL;
while(1)
{
printf("1.Add a shirt ");
printf("2.Sell a shirt ");
printf("3.Restock a shirt ");
printf("4.Print a list of all shirts ");
printf("5.Quit ");
printf("Enter your choice : ");
if(scanf("%d",&i)<=0){

printf("Enter only an Integer ");
exit(0);
} else {
switch(i)
{
case 1: printf("Enter the number to addaShirt : ");
scanf("%d",&nkkm);
addaShirt(nkkm);
break;
case 2: if(head==NULL)
{
printf("List is Empty ");
}
else
{
printf("Element(s) in the list are : ");
}
display(n);
break;
case 3: printf("Size of the list is %d ",count());
break;
case 4: if(head==NULL)
printf("List is Empty ");
else{
printf("Enter the number to delete : ");
scanf("%d",&nkkm);
if(delete(nkkm))
printf("%d deleted successfully ",nkkm);
else
printf("%d not found in the list ",nkkm);
}
break;
case 5: return 0;
default: printf("Invalid option ");
}
}

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Write a class named Deque20, placed in the file Deque20.java. It should behave exactly as described in the Java interface Deque (Google: \"Java 8 Deque\") except that it should only implement the following methods: a no-argument constructor, addFirst, addLast, removeFirst, removeLast, peekFirst, peekLast, and size. Your class should not extend or implement any class or interface. It must use an array as its storage, allow generic types (E), and be as efficient as possible. The Deque javadoc mentions several exceptions, but you should only return NullPointerException (if someone tries to add a null reference to the Deque) and NoSuchElementException (if someone tries to remove an item from an empty Deque). Your deque should store objects in a circular array. Probably the simplest implementation has objects with fields The meaning being that the elements currently in the deque are at positions list[first_index], list[(first_index+1)%list.length], list[(first_index+2)%list.length],..., list[(first_index+num_elems-1)%list.length]. The modulus is necessary because first_index might be near the end of the array and adding to it might result in an index that\'s out of bounds. When an add operation would cause the capacity of the array to be exceeded, you should double the size of the array by creating a new array of twice the size and copying the old array contents into the new. Do not worry about ever reducing the size of an array. Also, whenever a reference is removed from your deque, the old array location should be set to null so that the garbage collector knows that the old item is no longer referred to by the deque. Solution public class Deque20 { private int first_index; private int num_elems; private Object[] list; Deque20(){ first_index = 0; num_elems = 0; list = new Object[20]; } private void doubleSize(){ Object[] arr = new Object[2 * list.length]; for(int i = 0; i < list.length; ++i){ arr[i] = list[(i + first_index) % list.length]; } list = arr; } public void addFirst(E val){ if(list.length == num_elems){ doubleSize(); } int index = (first_index + list.length - 1) % list.length; list[index] = val; ++num_elems; } public void addLast(E val){ if(list.length == num_elems){ doubleSize(); } int index = (first_index + num_elems) % list.length; list[index] = val; ++num_elems; } public E removeFirst(){ int index = first_index; E ret = (E) list[index]; first_index = (first_index + 1) % list.length; return ret; } public E removeLast(){ int index = ((first_index + num_elems) % list.length) - 1; E ret = (E) list[index]; --num_elems; return ret; } public E peekFirst(){ return (E) list[first_index]; } public E peekLast(){ int index = ((first_index + num_elems) % list.length) - 1; return (E) list[index]; } public int size(){ return num_elems; } }.

Write a line of code to create a Print Writer to write to the file .pdf

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What is meant by ecological fallacy? Solution Answer: The term \"fallacy\" means the invalid. ecological fallacy tells about the individuals in the ecosystem which are invalid, it tell statistically about the group of individuals that are not part of the group which they should belong to. Eg. The earnings of people in particular district is high, the assumption that every person in that distict is rich would be fallacy. There are few people with low earnings..

what is the difference bettween cotranslational import and posttrans.pdf

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what is the difference bettween cotranslational import and posttranslational import of proteins into organelles? of the organelles listed below, which make use of which type of import? 1. ER 2. Mitochondria 3. nucleus 4. peroxisome Solution Cotranslational transport involves translocation of a growing peptide chain within the cell simultaneously like to the ER. Where as, post translational transport involves initially protein synthesis in the cytosol followed by transport to ER. Cotranslational - ER, Post translational- Mitochondria, nucleus and peroxisome.

What is the meaning of critical periods in neonatal development Wha.pdf

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what are some necessary organizational steps to successfully offer acupressure as part of nursing care in the labor and delivery setting? Solution Principles of acupuncture forms the basis of acupressure. During labour it can be used to relieve pain and increase the efficiency of the labor. From 37th week onwards all points are safe for acupressure. Lesser pressure as compared to the nirmal times has to be applied during labor. It should be started during early stages of labor and shall intermittently be given during the entire process. There are few specific acupressure points during the delivery and labor. These are lower back points, large intestine, urinary bladder, kidney and spleen..

What are the major differences between foreign bonds and Eurobonds .pdf

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What is Incorporation by Reference? and How is it used in a contract? Solution Incorporation by reference is an act of including second document within the first document by giving only reference in first document. Ie if in first document only a single line is kept giving reference of second document. than the secound document will be part of first document. Incorporation by reference is often done in creating law as well in contract law nad trust and estate law Incorporation by reference is used in contract by way of terms and condition. For example there are general set of terms and condition to contarct now instead of including all the tems and condtion to contract a company can just give reference to it, making it binding to the companies entering the contract. Thus it is very important to read document if it is included by reference.

What is meant by ecological fallacySolutionAnswerThe term .pdf

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What are the major differences between foreign bonds and Eurobonds? Why do some US companies prefer to issue Eurobonds to raise US dollars from foreign markets? Solution A Eurobond is long term bond issued and sold by the compnies outside the country in which it is dominated. If US company issued a bond in dollar and selling it to Japanese investors in Japan then it is know as Eurobond. Foreign is also sold in the other courtry but it is issued in that country currency as well that\'s a difference from Eurobond. US company selling and issuing a bond in Yen for Japanese investors. US companies issue Eurobonds because in that way US companies can hedge its currency risk..

What are the two main branches of Non-Euclidean Geometry What is the.pdf

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By what transport method does oxygen enter the blood from the alveoli? By what method is oxygen transported to the body tissues from the lungs? By what method does oxygen enter the body tissues from the blood? a. Diffusion, bulk flow, diffusion b. Bulk flow, bulk flow, diffusion c. Bulk flow, diffusion, carrier proteins d. Diffusion, diffusion, carrier proteins e. Bulk flow, diffusion, bulk flow Solution b. Bulk flow, bulk flow, diffusion bulk flow carries air or water, relatively high in O2 and low in CO2, past a respiratory surface. The flow is usually propelled by muscular movements. Bulk flow of blood transports gases between the respiratory system and the tissues. Blood is pumped through the body by the heart. Diffusion transfers O2 out of the capillaries and into nearby tissues, and transfers CO2 from the tissues into capillaries.

By what transport method does oxygen enter the blood from the alveol.pdf

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All of these are examples of evidence used to support the hypothesis of endosymbiosis, EXCEPT:a. Mitochondria and plastids contain circular DNA molecules similar to the circular chromosomes of bacteriab. Mitochondria and plastids, like prokaryotes, only contain genetic material in the form of RNAc. Mitochondria and plastids can replicate themselves in a process similar to prokaryotesd. Mitochondria and plastids have their own ribosomes and enzymes that can be used to transcribe and translate DNA into proteinse. Mitochondria and plastids have a double membrane (inner and outer) Solution Out of the above options , the correct will be B. Mitochondria and plastids a have their genetic material in the form of DNA not RNA. They donot have any such machinary for their expression , so this option is incorrect regarding endosymbiosis. Otherwise rest all the sentenses are true. They have circular genome like Prokaryotes, and can replicate and also have double membrane layer..

All of these are examples of evidence used to support the hypothesis.pdf

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A 28 year-old biologist makes a trip to study life forms in a distant planet 17 light-years away. The round trip including a stay of 1 year in the distant planet takes 38 years according to the clock on earth. The biologist\'s son is 2.0 year-old when she left. Assume that the planet is not moving with respect to earth. How fast is the spaceship traveling? Give your answer in terms of the speed of light. You do not need to enter \"c\" as unit. How old is she when she returns according to her clock?How old is her son according to the clock on earth? What is the total distance traveled according to the biologist? Give your answer in light-years. You do not need to enter the units. Solution As it takes 38 - 1 = 37 years for the biologist to make the trip Speed = 17 / 38 = 0.45 c Time dialation that is occurred t\' = t * sqrt(1 - (v2/c2)) t\' = 37 * sqrt(1 - 0.452) = 33.04 years Biologists age = 28 + 33.04 + 1 = 62.04 years Her son\'s age = 2.0 + 38 = 40 years.

A 28 year-old biologist makes a trip to study life forms in a distan.pdf

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4. Name the following compounds: CO2 CH CH3 CH20H CH3COOH 5. Water has a molecular weight of 18. A milliliter of liquid water has a mass of one gram. Calculate the molar concentration of liquid water: 6. What kind of bonds hold together the following compounds (match): hydrogen bonds water ionic bonds table salt covalent bonds nucleotides in RNA james bonds strands in dsDNA 7. Fill in the blanks: is transcribed by to make is translated by to make Solution Answer: 4) CO2 : carbondioxide O2: Oxygen N2: NItrogen CH4 : Methane CH3CH2OH : Ethyl alcohol CH3COOH: Acetic acid.

4. Name the following compounds CO2 CH CH3 CH20H CH3COOH 5. Water ha.pdf

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4. When examining whether group differences occur on more than one d.pdf

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