未能推断出模板参数,求调试成功!

[majiayefengkuanghaha]

#include <iostream>
#include <vector>
#include <functional>
#include <string>

using namespace std;

class A
{
public:
string name;
void print(ostream& out) const
{
out << name << endl;
}
};

ostream& operator<<(ostream& out, const A& a)
{
a.print(out);
return out;
}

template <typename T, typename T1>
const T& findMax(const vector<T> & a, T1 b = less<T1>)
{
int maxIndex = 0;
for(int i = 1; i != a.size(); ++i)
{
if(b(a[maxIndex], a[i]))
{
maxIndex = i;
}
}
return a[maxIndex];
}

class B
{
public:
bool operator()(const string& a, const string& b) const
{
return _stricmp(a.c_str(), b.c_str()) < 0;
}
};

int main()
{
vector<A> a(3);
a[0].name = "A";
a[1].name = "a";
a[2].name = "c";
cout << findMax(a) << endl;
cout << findMax(a,B()) << endl;
}

[HeartWasNot]

用容器的模板是。。。貌似要用typename vector<T> a如此这般。。。。


PS：LZ该结贴了。。你这结贴率。。。没人愿意回答你的问题。。。

[majiayefengkuanghaha]

好的,我来结贴!

[majiayefengkuanghaha]

晕死.原来是元素问题...

[qq120848369]

#include <iostream>
#include <vector>
#include <functional>
#include <string>

using namespace std;

class A
{
public:
	string name;
	void print(ostream& out) const
	{
		out << name << endl;
	}

	bool operator < (const A &a) const	//
	{
		return name<a.name;
	}
};

ostream& operator<<(ostream& out, const A& a)
{
	a.print(out);
	return out;
}

template <typename T,typename T1>
const T& findMax(const vector<T> & a, T1 b)//
{
	int maxIndex = 0;
	for(int i = 1; i != a.size(); ++i)
	{
		if(b(a[maxIndex], a[i]))
		{
			maxIndex = i;
		}
	}
	return a[maxIndex];
}

template <typename T>
const T& findMax(const vector<T> & a)//
{
	int maxIndex = 0;
	less<T> b;

	for(int i = 1; i != a.size(); ++i)
	{
		if(b(a[maxIndex], a[i]))
		{
			maxIndex = i;
		}
	}
	return a[maxIndex];
}

class B
{
public:
	bool operator()(const A& a, const A& b) const
	{
		return a.name<b.name ;
	}
};

int main()
{
	vector<A> a(3);
	a[0].name = "A";
	a[1].name = "a";
	a[2].name = "c";
	cout << findMax(a) << endl;
	cout << findMax(a,B()) << endl;  
	
	return 0;
}

模板函数没有特化，没有默认模板参数，只能重载，而且默认less的话需要提供A的<。