STL源代码剖析 容器 stl_stack.h

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stack

-------------------------------------------------------------
stack 是一种配接器(adapter)。以某种容器作为底部结构,改变其接口,使之符合"先进后出"的特性。
SGI STL 默认以 deque 为 stack 底部结构
没有遍历行为,没有遍历器
演示样例:
#include <stack>
#include <list>
#include <iostream>
#include <algorithm>
using namespace std;


int main(){
	stack<int, list<int> >istack;
	istack.push(1);
	istack.push(2);
	cout << istack.size() << endl
		 << istack.top() << endl;
}

源代码:
#ifndef __SGI_STL_INTERNAL_STACK_H
#define __SGI_STL_INTERNAL_STACK_H


__STL_BEGIN_NAMESPACE


#ifndef __STL_LIMITED_DEFAULT_TEMPLATES
template <class T, class Sequence = deque<T> > //默认以 deque 为底层容器
#else
template <class T, class Sequence>
#endif
class stack {
  friend bool operator== __STL_NULL_TMPL_ARGS (const stack&, const stack&);
  friend bool operator< __STL_NULL_TMPL_ARGS (const stack&, const stack&);
public:
  typedef typename Sequence::value_type value_type;
  typedef typename Sequence::size_type size_type;
  typedef typename Sequence::reference reference;
  typedef typename Sequence::const_reference const_reference;
protected:
  Sequence c; //底层容器
public:
  //下面全然利用 Sequence c 的操作完毕 stack 的操作
  bool empty() const { return c.empty(); }
  size_type size() const { return c.size(); }
  reference top() { return c.back(); }
  const_reference top() const { return c.back(); }
  //改动接口使符合 stack "前进后出"的特性
  void push(const value_type& x) { c.push_back(x); }
  void pop() { c.pop_back(); }
};


template <class T, class Sequence>
bool operator==(const stack<T, Sequence>& x, const stack<T, Sequence>& y) {
  return x.c == y.c;
}


template <class T, class Sequence>
bool operator<(const stack<T, Sequence>& x, const stack<T, Sequence>& y) {
  return x.c < y.c;
}


__STL_END_NAMESPACE


#endif /* __SGI_STL_INTERNAL_STACK_H */


// Local Variables:
// mode:C++
// End:


posted @ 2017-07-01 19:12  wzjhoutai  阅读(276)  评论(0编辑  收藏  举报