STL-------set与multiset

1.set(集）、multiset(多集）

2.红黑树

3.基本操作

          insert       

          count和find

          erase

注意：不能通过find进行修改

#include <iostream>
#include <set>

using namespace std;

template <typename Container>
void PrintContents(const Container& c)
{
	//必须加typename  否则linux下编译是不通过的
	typename Container::const_iterator i = c.begin();
	
	while( i!=c.end() )
	{
		cout<<*i<<endl;
		++i;
	}
}

int main( int argc, char** argv )
{
	set<int> a;
	multiset<int> ma;

	a.insert(60);
	a.insert(-1);
	a.insert(3000);
	PrintContents(a);
	/*
	set<int>::const_iterator citr = a.begin();
	for( citr; citr!=a.end(); ++citr )
	{
		cout<<*citr<<endl;
	}*/

	ma.insert(a.begin(), a.end());
	ma.insert(3000);
	ma.insert(3000);
	cout<<"3000 --- total:"<<ma.count(3000)<<endl;
	PrintContents(ma);
	/*
	multiset<int>::const_iterator itr = ma.begin();
	for( ; itr!=ma.end(); ++itr )
	{
		cout<<*itr<<endl;
	}*/

	return 0;
}

请注意 typename Container::const_iterator i ;一定要加上typename。即使有些编译器让你通过。

不写的话可能有以下错误

set.cpp: In function `void PrintContents(const Container&)':
set.cpp:9: error: expected `;' before "i"
set.cpp:11: error: `i' undeclared (first use this function)
set.cpp:11: error: (Each undeclared identifier is reported only once for each function it appears
.)
set.cpp: In function `void PrintContents(const Container&) [with Container = std::set<int, std::le
<int>, std::allocator<int> >]':
set.cpp:26:   instantiated from here
set.cpp:9: error: dependent-name ` Container::const_iterator' is parsed as a non-type, but instant
tion yields a type
set.cpp:9: note: say `typename  Container::const_iterator' if a type is meant
set.cpp: In function `void PrintContents(const Container&) [with Container = std::multiset<int, st
:less<int>, std::allocator<int> >]':
set.cpp:38:   instantiated from here
set.cpp:9: error: dependent-name ` Container::const_iterator' is parsed as a non-type, but instant
tion yields a type
set.cpp:9: note: say `typename  Container::const_iterator' if a type is meant

解释：It accurately doesn't treat map<T, A>::const_iterator as a type because it relies on template parameters, T and A. To make the compiler believe you, you need to use the typename keyword.