一个类可以有多个拷贝构造函数，那么怎么决定调用哪个复制构造函数呢？

[愚蠢的萝卜]

C++初学
在书上看到：一个类至少有 一个拷贝构造函数；事实上，可以有多个。
而且刚刚看到以下代码：

class ROSCPP_DECL NodeHandle
  {
  public:
    /**
     * \brief Constructor
     *
     * When a NodeHandle is constructed, it checks to see if the global
     * node state has already been started.  If so, it increments a
     * global reference count.  If not, it starts the node with
     * ros::start() and sets the reference count to 1.
     *
     * \param ns Namespace for this NodeHandle.  This acts in addition to any namespace assigned to this ROS node.
     *           eg. If the node's namespace is "/a" and the namespace passed in here is "b", all 
     *           topics/services/parameters will be prefixed with "/a/b/"
     * \param remappings Remappings for this NodeHandle.
     * \throws InvalidNameException if the namespace is not a valid graph resource name
     */
    NodeHandle(const std::string& ns = std::string(), const M_string& remappings = M_string());
    /**
     * \brief Copy constructor
     *
     * When a NodeHandle is copied, it inherits the namespace of the
     * NodeHandle being copied, and increments the reference count of
     * the global node state by 1.
     */
    NodeHandle(const NodeHandle& rhs);
    /**
     * \brief Parent constructor
     *
     * This version of the constructor takes a "parent" NodeHandle, and is equivalent to:
     \verbatim
     NodeHandle child(parent.getNamespace() + "/" + ns);
     \endverbatim
     *
     * When a NodeHandle is copied, it inherits the namespace of the
     * NodeHandle being copied, and increments the reference count of
     * the global node state by 1.
     *
     * \throws InvalidNameException if the namespace is not a valid
     * graph resource name
     */
    NodeHandle(const NodeHandle& parent, const std::string& ns);
    /**
     * \brief Parent constructor
     *
     * This version of the constructor takes a "parent" NodeHandle, and is equivalent to:
     \verbatim
     NodeHandle child(parent.getNamespace() + "/" + ns, remappings);
     \endverbatim
     *
     * This version also lets you pass in name remappings that are specific to this NodeHandle
     *
     * When a NodeHandle is copied, it inherits the namespace of the NodeHandle being copied, 
     * and increments the reference count of the global node state
     * by 1.
     * \throws InvalidNameException if the namespace is not a valid graph resource name
     */
    NodeHandle(const NodeHandle& parent, const std::string& ns, const M_string& remappings);
    /**
     * \brief Destructor
     *
     * When a NodeHandle is destroyed, it decrements a global reference
     * count by 1, and if the reference count is now 0, shuts down the
     * node.
     */

NodeHandle(const NodeHandle& rhs)

显然是NodeHandle的拷贝构造函数，但是

NodeHandle(const NodeHandle& parent, const std::string& ns)， NodeHandle(const NodeHandle& parent, const std::string& ns, const M_string& remappings)

这两个从注释来看，应该也是起拷贝作用。所以就在想如果一个类有多个拷贝构造函数，那么编译器是怎么判断调用哪个拷贝构造函数的呢？

[愚蠢的萝卜]

引用 1 楼 u010165006 的回复:

根据调用（构造）时的函数实参

谢谢

[ooolinux]

根据调用（构造）时的函数实参

[珍惜生命远离CPP]

NodeHandle(const NodeHandle& rhs); 拷贝只有这1个 其他的都是构造函数