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#2
玩出来的代码2010-11-28 14:40
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#include <iostream>
#include <list>
using namespace std;
class Test
{
public:
Test(int a=0) { this->data_ = a; }
~Test() {}
int data() {return this->data_;}
void data(int a) { this->data_ = a; }
bool operator < (Test* right) const
{
return this->data_ <= right->data();
}
private:
int data_;
};
typedef std::list<Test*> TestList;
template<>
struct std::greater<Test*>
{
bool operator()( Test*_X, Test* _Y) const
{
return (_X->data() > _Y->data());
}
};
int main()
{
TestList a;
int i = 0;
Test *p[5];
p[0] = new Test(65);
p[1] = new Test(34);
p[2] = new Test(45);
p[3] = new Test(23);
p[4] = new Test(28);
a.push_back(p[0]);
a.push_back(p[1]);
a.push_back(p[2]);
a.push_back(p[3]);
a.push_back(p[4]);
a.sort(greater<Test*>());
TestList::iterator iter = a.begin();
for(; iter != a.end(); ++iter)
cout<<(*iter)->data() << " ";
cout <<endl;
return 0;
}
为什么上述的STL定义体中的template<>没有参数呢 ?#include <list>
using namespace std;
class Test
{
public:
Test(int a=0) { this->data_ = a; }
~Test() {}
int data() {return this->data_;}
void data(int a) { this->data_ = a; }
bool operator < (Test* right) const
{
return this->data_ <= right->data();
}
private:
int data_;
};
typedef std::list<Test*> TestList;
template<>
struct std::greater<Test*>
{
bool operator()( Test*_X, Test* _Y) const
{
return (_X->data() > _Y->data());
}
};
int main()
{
TestList a;
int i = 0;
Test *p[5];
p[0] = new Test(65);
p[1] = new Test(34);
p[2] = new Test(45);
p[3] = new Test(23);
p[4] = new Test(28);
a.push_back(p[0]);
a.push_back(p[1]);
a.push_back(p[2]);
a.push_back(p[3]);
a.push_back(p[4]);
a.sort(greater<Test*>());
TestList::iterator iter = a.begin();
for(; iter != a.end(); ++iter)
cout<<(*iter)->data() << " ";
cout <<endl;
return 0;
}
template<>
struct std::greater<Test*>
{
bool operator()( Test*_X, Test* _Y) const
{
return (_X->data() > _Y->data());
}
};