转换用户数据类型

这是例如由教程从here：有这样类：转换用户数据类型

struct Vec3 { double x, y, z; bool operator==(Vec3& other) const { return other.x == this->x && other.y == this->y && other.z == this->z;}};

此代码为YAML数据转换为用户对象（Vec3）：

namespace YAML { 
template<> 
struct convert<Vec3> { 
    static bool decode(const Node& node, Vec3& rhs) { 
    if(!node.IsSequence() || node.size() != 3) { 
     return false; 
    } 
    rhs.x = node[0].as<double>(); 
    rhs.y = node[1].as<double>(); 
    rhs.z = node[2].as<double>(); 
    return true; 
    } 
}; 
}

现在我测试一下：

YAML::Node node = YAML::Load("start: [1, 3, 0]"); 
Vec3 v = node["start"].as<Vec3>();

但我有一个错误：

In file included from /usr/local/include/yaml-cpp/yaml.h:17:0, 
       from /home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:5: 
/usr/local/include/yaml-cpp/node/impl.h: In instantiation of ‘T YAML::as_if<T, void>::operator()() const [with T = Vec3]’: 
/usr/local/include/yaml-cpp/node/impl.h:146:31: required from ‘T YAML::Node::as() const [with T = Vec3]’ 
/home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:52:37: required from here 
/usr/local/include/yaml-cpp/node/impl.h:122:7: error: no matching function for call to ‘Vec3::Vec3()’ 
    T t; 
    ^
/home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:14:5: note: candidate: Vec3::Vec3(double, double, double) 
    Vec3(double a, double b, double c){ 
    ^
/home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:14:5: note: candidate expects 3 arguments, 0 provided 
/home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:12:8: note: candidate: constexpr Vec3::Vec3(const Vec3&) 
struct Vec3 { 
     ^
/home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:12:8: note: candidate expects 1 argument, 0 provided 
/home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:12:8: note: candidate: constexpr Vec3::Vec3(Vec3&&) 
/home/ken/ClionProjects/First/yaml-cpp/first_yaml.cpp:12:8: note: candidate expects 1 argument, 0 provided 
CMakeFiles/First.dir/build.make:62: recipe for target 'CMakeFiles/First.dir/yaml-cpp/first_yaml.cpp.o' failed

如何正确转换数据类型？

来源

2016-12-01 Roma Karageorgievich

Vec3是你实际使用的定义吗？该错误消息表明您需要Vec3的默认构造函数，并且它只发现Vec3(double, double, double)（加上复制/移动）。

如果您定义了一个需要三个双打的构造函数，那么您还需要定义一个接受零参数的构造函数。

来源

2016-12-01 15:07:36

转换用户数据类型

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