// boost.lambda表达式用法
// made by davidsu33
// 2014-9-22

#include "stdafx.h"

#include <boost/typeof/typeof.hpp>
#include <boost/lambda/lambda.hpp>
#include <boost/bind/bind.hpp>
#include <boost/assign.hpp>
#include <boost/core/enable_if.hpp>

#include <vector>
#include <utility>
#include <memory>
#include <algorithm>
#include <cassert>
#include <iostream>

using namespace std;
//Boost的简单用法
void lambda_easy_use()
{
	using namespace boost::assign;
	vector<int> arr = list_of(0).repeat(19, 0);

	int begin = 0;
	for_each(arr.begin(), arr.end(),
		(
		boost::lambda::_1 = boost::lambda::var(begin)++, 
		boost::lambda::_1 += 10,
		boost::lambda::_1 -= 5
		)
		);

	for (unsigned i=0; i<arr.size(); ++i)
	{
		assert(arr[i] == 5+i);
	}

	cout<<"lambda_easy_use passed"<<endl;
}

//Boost的仿函数调用
void lambda_functors()
{
	int i=10, j=20;
	int r = (boost::lambda::_1 + boost::lambda::_2)(i,j);
	assert(r == (i+j));

	BOOST_AUTO(br, (boost::lambda::_1 - boost::lambda::_2)(i, j));
	assert(br == (i-j));

	//多次运算
	BOOST_AUTO(br2, 
		(boost::lambda::_1 - boost::lambda::_2, boost::lambda::_1 += boost::lambda::constant(10))(i, j));

	//??br2 == 20，br2的取值由第二个表达式赋予
	assert(br2 == i);
	assert(i == 20);
}

template<class T>
T fun(const T& value)
{
	//要求输入值必须为整型
	typedef boost::enable_if<boost::is_integral<T>, T>::type TType;

	cout<<"fun is called!"<<endl;
	cout<<"the value:"<<value<<endl;
	return value+100;
}

void lambda_call()
{
	vector<int> v;
	using namespace boost::assign;
	v += 1,2,4,8,16;
	for_each(v.begin(), v.end(), fun<int>);
}

int _tmain(int argc, _TCHAR* argv[])
{
	lambda_easy_use();
	lambda_functors();


	getchar();
	return 0;
}

boost.lambda