C++ 错误 C2678：二进制“==”：未找到采用左侧操作数类型的运算符（或没有可接受的转换）
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error C2678: binary '==' : no operator found which takes a left-hand operand of type (or there is no acceptable conversion)
提问by Andrey Dyatlov
I'm trying to compile the following code:
我正在尝试编译以下代码：
#include <boost/geometry/geometries/point_xy.hpp>
#include <iostream>
#include <utility>
typedef boost::geometry::model::d2::point_xy<long> Point;
typedef std::pair<Point, Point> Vector;
bool operator==(const Point& p1, const Point& p2) {
return p1.x() == p2.x() && p1.y() == p2.y();
}
int main() {
Vector vec1(Point(0,0), Point(1,1));
Vector vec2(Point(0,0), Point(1,2));
std::cout << ((vec1 == vec2) == false) << std::endl;
std::cout << ((vec1 == vec1) == true) << std::endl;
}
VS2012 C++ compiler returns the following compilation error:
VS2012 C++编译器返回如下编译错误：
...VC\include\utility(219): error C2678: binary '==' : no operator found which takes a left-hand operand of type 'const Point' (or there is no acceptable conversion)
...VC\include\utility(219)：错误 C2678：二进制“==”：未找到采用“const Point”类型左侧操作数的运算符（或没有可接受的转换）
GCC C++ compiler returns the following compilation error:
GCC C++ 编译器返回以下编译错误：
/usr/include/c++/4.8/bits/stl_pair.h:
In instantiation of ‘bool std::operator==(const std::pair<_T1, _T2>&, const std::pair<_T1, _T2>&) [with _T1 = boost::geometry::model::d2::point_xy; _T2 = boost::geometry::model::d2::point_xy]':
test.cpp:22:28: required from here /usr/include/c++/4.8/bits/stl_pair.h:215:51: error:
no match for ‘operator==' (operand types are ‘const boost::geometry::model::d2::point_xy' and ‘const boost::geometry::model::d2::point_xy') { return __x.first == __y.first && __x.second == __y.second; }
/usr/include/c++/4.8/bits/stl_pair.h：
在实例化 'bool std::operator==(const std::pair<_T1, _T2>&, const std::pair<_T1, _T2>&) [with _T1 = boost::geometry::model::d2 ::point_xy; _T2 = boost::geometry::model::d2::point_xy]':
test.cpp:22:28: 从这里需要 /usr/include/c++/4.8/bits/stl_pair.h:215:51: 错误：
'operator==' 不匹配（操作数类型是 'const boost::geometry::model::d2::point_xy' 和 'const boost::geometry::model::d2::point_xy'）{ return __x.第一个 == __y.first && __x.second == __y.second; }
Error disappears if I overload == operator for Vector:
如果我为 Vector 重载 == 运算符，错误就会消失：
bool operator==(const Vector& v1, const Vector& v2) {
return v1.first == v2.first && v1.second == v2.second;
}
回答by Angew is no longer proud of SO
The reason why this fails is that the operator ==for std::pairuses ==to compare the pairs' members, which in turn uses argument-dependent lookup (ADL)to find the proper operator ==for them. But you've provided the overload in the wrong namespace, since Pointis actually a typedef for something in ::boost::geometry::model::d2, and not in ::.
失败的原因是operator ==forstd::pair用于==比较对的成员，而后者又使用依赖于参数的查找 (ADL)来找到适合operator ==它们的成员。但您提供在错误的命名空间中的过载，因为Point实际上是在东西的typedef ::boost::geometry::model::d2，而不是在::。
If you move the operator into the correct namespace (which is a good idea anyway), it works:
如果您将运算符移动到正确的命名空间（无论如何这是个好主意），它就可以工作：
#include <boost/geometry/geometries/point_xy.hpp>
#include <iostream>
#include <utility>
typedef boost::geometry::model::d2::point_xy<long> Point;
typedef std::pair<Point, Point> Vector;
namespace boost { namespace geometry { namespace model { namespace d2 {
bool operator==(const Point& p1, const Point& p2) {
return p1.x() == p2.x() && p1.y() == p2.y();
}
} } } }
int main() {
Vector vec1(Point(0,0), Point(1,1));
Vector vec2(Point(0,0), Point(1,2));
std::cout << ((vec1 == vec2) == false) << std::endl;
std::cout << ((vec1 == vec1) == true) << std::endl;
}
