您的位置:首页 > 编程语言 > C语言/C++

Vector的简单C++实现

2014-06-29 22:52 295 查看
template <typename Object>
class Vector
{
public:
explicit Vector(int initSize=0)
:theSize( initSize ),theCapacity( initSize + SPARE_CAPACITY)
{objects = new Object[theCapacity];}

Vector(const Vector & rhs): objects(NULL)
{ operator= (rhs);}

~Vector(void)
{delete[] objects;}

const Vector &operator= (const Vector & rhs)
{
if(this != &rhs)
{
delete[]objects;
theSize = rhs.size();
theCapacity = rhs.theCapacity;

objects = new Object[capacity()];
for(int k=0;k<size();k++)
objects[k] = rhs.objects[k];
}
return *this;
}

void resize(int newSize)
{
if(newSize > theCapacity)
{
reserve(newSize*2+1);
theSize = newSize;
}
}

void reserve(int newCapacity)
{
if(newCapacity < theSize)
return;

Object *oldArray = objects;

objects = new Object[newCapacity];
for(int k=0;k<theSize;k++)
objects[k] = oldArray[k];

theCapacity = newCapacity;

delete [] oldArray;
}

Object & operator[](int index)
{ return objects[index];}
const Object & operator[](int index) const
{ return objects[index];}

bool empty() const
{ return size() == 0;}
int size() const
{ return theSize;}
int capacity() const
{ return theCapacity;}

void push_back( const Object & x)
{
if(theSize == theCapacity)
reserve(2*theCapacity + 1);
objects[theSize++] = x;
}

void pop_back()
{ theSize--;}

const Object & back() const
{ return objects[theSize - 1];}

typedef Object * iterator;
typedef const Object * const_iterator;

iterator begin()
{ return &objects[0];}
const_iterator begin() const
{ return &objects[0];}
iterator end()
{ return &objects[size()];}
const_iterator end() const
{ return &objects[size()];}

enum{SPARE_CAPACITY = 16};

private:
int theSize;
int theCapacity;
Object * objects;
};
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: