#include <iostream>
#include <vector>
#include <iterator>
using namespace std;
struct Point
{
int x;
int y;
Point(int x, int y) :
x(x),
y(y)
{}
};
int main()
{
vector<Point> points;
points.push_back(Point(1, 2));
points.push_back(Point(4, 6));
vector<int> xs;
for(vector<Point>::iterator it = points.begin();
it != points.end();
++it)
{
xs.push_back(it->x);
}
copy(xs.begin(), xs.end(), ostream_iterator<int>(cout, " "));
cout << endl;
return 0;
}
我想知道如何使用STL算法获得与上面的for循环相同的结果?我已经尝试了一些使用for_each的东西,但是无法让它工作。
答案 0 :(得分:7)
您不会使用std::for_each
,而是std::transform
(您正在将某个点转换为单个数字。)
例如:
#include <algorithm> // transform resides here
#include <iostream>
#include <iterator>
#include <vector>
struct Point
{
int x;
int y;
Point(int x, int y) :
x(x),
y(y)
{
}
};
int point_to_int(const Point& p)
{
return p.x;
}
int main()
{
std::vector<Point> points;
points.push_back(Point(1, 2));
points.push_back(Point(4, 6));
std::vector<int> xs;
std::transform(points.begin(), points.end(),
std::back_inserter(xs), point_to_int);
std::copy(xs.begin(), xs.end(),
std::ostream_iterator<int>(std::cout, " "));
std::cout << std::endl;
return 0;
}
因为您知道要转换的容器的大小,所以您可能会从以下方面获得轻微的性能提升。我也发现它更具可读性:
std::vector<int> xs;
xs.reserve(points.size());
std::transform(points.begin(), points.end(),
std::back_inserter(xs), point_to_int);
#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
#include <boost/bind.hpp>
#include <boost/lambda/lambda.hpp>
struct Point
{
int x;
int y;
Point(int x, int y) :
x(x),
y(y)
{
}
};
int main()
{
using namespace boost;
std::vector<Point> points;
points.push_back(Point(1, 2));
points.push_back(Point(4, 6));
std::vector<int> xs;
xs.reserve(points.size());
std::transform(points.begin(), points.end(),
std::back_inserter(xs), bind(&Point::x, lambda::_1));
std::copy(xs.begin(), xs.end(),
std::ostream_iterator<int>(std::cout, " "));
std::cout << std::endl;
return 0;
}
无需在其他地方指定功能。这使代码保持靠近调用站点,并且通常可以提高可读性。
在C ++ 0x中,它只是:
std::transform(points.begin(), points.end(),
std::back_inserter(xs), [](const Point& p){ return p.x; } );
(据我所知,无论如何)