将数组值解压缩为Cpp中的类变量

Question

我想要将数组值解压缩到不同的类变量中，但是我得到了一个错误。

auto [SendLowROS.motorCmd[FR_0].Kp, SendLowROS.motorCmd[FR_1].Kp, SendLowROS.motorCmd[FR_2].Kp, 
      SendLowROS.motorCmd[FL_0].Kp, SendLowROS.motorCmd[FL_1].Kp, SendLowROS.motorCmd[FL_2].Kp, 
      SendLowROS.motorCmd[RR_0].Kp, SendLowROS.motorCmd[RR_1].Kp, SendLowROS.motorCmd[RR_2].Kp, 
      SendLowROS.motorCmd[RL_0].Kp, SendLowROS.motorCmd[RL_1].Kp, SendLowROS.motorCmd[RL_2].Kp] = msg.Kp;

/home/src/llm.cpp: In member function ‘void Driver::jointCommandCallback(msgs::JointCMD)’:
/home/src/llm.cpp:65:25: error: expected ‘]’ before ‘.’ token
         auto [SendLowROS.motorCmd[FR_0].Kp, SendLowROS.motorCmd[FR_1].Kp, SendLowROS.motorCmd[FR_2].Kp,
                         ^
/home/src/mbs_unitree_ros/src/llm.cpp:68:111: error: ‘std::_Vector_base<float, std::allocator<float> >’ is an inaccessible base of ‘std::vector<float>’
               SendLowROS.motorCmd[RL_0].Kp, SendLowROS.motorCmd[RL_1].Kp, SendLowROS.motorCmd[RL_2].Kp] = msg.Kp;

Answer 1

你不能这么做。

结构化绑定只是为您要解包的任何部分创建新的唯一名称。

例如：

auto&& [a,b,c] = std::tuple{1,2,3};
some_class.a = a;
some_class.b = b;
some_class.c = c;

详情请参见https://en.cppreference.com/w/cpp/language/structured_binding。

Answer 2

您不能绑定到结构化绑定中的现有变量。

为此，您可以改用std::tie()。

示例：

#include <iostream>
#include <tuple>

int main() {
    int a, b, c
    std::tie(a, b, c) = std::tuple(1, 2, 3);
    std::cout << a << b << c;                 // prints 123
}

或者对于不可复制但可移动的类型：

#include <iostream>
#include <tuple>

struct foo {
    foo() = default;
    foo(const foo&) = delete;            // not copyable for the sake of this demo
    foo(foo&&) = default;                // but moveable
    foo& operator=(const foo&) = delete;
    foo& operator=(foo&&) = default;

    int a;
};

int main() {
    foo w, x, y{3}, z{4};
    
    std::tie(w, x) = std::tuple(std::move(y), std::move(z));

    std::cout << w.a << x.a << '\n';     // prints 34
}