explain what the program due in letters. #include using namespace std; class Vec { public: Vec() { arr = new int[1]; this->cap= 1;//we start with cap=1 then we reserve to cap =10 ,and then when sz=cap we double the cap this->sz= 0; } int size() { return this->sz; } int capacity() { return this->cap; } void reserve( int n )//me:here we creat new arr with double capacity and move the old array to this new cap { // TODO:
explain what the program due in letters. #include using namespace std; class Vec { public: Vec() { arr = new int[1]; this->cap= 1;//we start with cap=1 then we reserve to cap =10 ,and then when sz=cap we double the cap this->sz= 0; } int size() { return this->sz; } int capacity() { return this->cap; } void reserve( int n )//me:here we creat new arr with double capacity and move the old array to this new cap { // TODO:
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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OOPs
In today's technology-driven world, computer programming skills are in high demand. The object-oriented programming (OOP) approach is very much useful while designing and maintaining software programs. Object-oriented programming (OOP) is a basic programming paradigm that almost every developer has used at some stage in their career.
Constructor
The easiest way to think of a constructor in object-oriented programming (OOP) languages is:
Question
explain what the program due in letters.
#include <iostream>
using namespace std;
class Vec
{
public:
Vec()
{
arr = new int[1];
this->cap= 1;//we start with cap=1 then we reserve to cap =10 ,and then when sz=cap we double the cap
this->sz= 0;
}
int size()
{
return this->sz;
}
int capacity()
{
return this->cap;
}
void reserve( int n )//me:here we creat new arr with double capacity and move the old array to this new cap
{
// TODO:
// (0) check the n should be > size, otherwise
// ignore this action.
if ( n > sz )
{
// (1) create a new int array which size is n
// and get its address
int *newarr = newint[n];
// (2) use for loop to copy the old array to the
// new array
int i;
for(i=0;i<sz;i++)
{
newarr[i] = this->arr[i];
}
// (3) update the variable to the new address
//me: to remember the old arr
int* oldarr = this->arr;//by that we copy they old arr to the new arry and we still able to remeber the old arr
this->arr = newarr;
// (4) delete old array
delete[] oldarr;
cap = n;
arr = newarr;
}
}
void push_back( int v )
{
// TODO:
//
if ( sz == cap )
{
cap *= 2; //here we double the cap
reserve(cap);
}
// complete others
this->arr[sz] = v;
sz++;
}
int at( int idx )//read the arr
{
returnthis->arr[idx];
}
private:
int *arr;
int sz = 0;
int cap = 0;
};
int main()
{
Vec v;
v.reserve(10);//the new cap, so here we creat a new capacite =10 then when the size becomes equ to cap it will double the cap to 20
v.push_back(3);
v.push_back(2);
cout << v.size() << endl; // 2
cout << v.capacity() << endl; // 10
v.push_back(3);
v.push_back(2);
v.push_back(3);
v.push_back(4);
v.push_back(3);
v.push_back(7);
v.push_back(3);
v.push_back(8);
v.push_back(2);
cout << v.size() << endl; // 11
//when the size becomes equal to the cap the cap doubles up
cout << v.capacity() << endl; // 20
for ( int i = 0; i < v.size(); i++ )
{
cout << v.at(i) << endl;
}
return 0;
}
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