Question

为什么这里声明的对象没有被移动，而是被复制了？它与拷贝省略有什么关系吗？目前的代码不应该改变班级＆＃39;对象？最初预计通过使用std :: move，B的内部结构将会发生变化。我忽略了什么吗？有没有解决方法？

#include <iostream>
#include <algorithm>
#include <vector>

class A {
    public:
        A() {}
        auto&& get() { return v; }
        friend std::ostream& operator<<(std::ostream& o, A const& a) {
            o << "A: ";
            for (auto it = std::begin(a.v); it != std::end(a.v); ++it) { if (it != std::begin(a.v)) { o << " "; } o << *it; }
            return o;
        }
    private:
        std::vector<int> v;
};

class B {
    public:
        B() {};
        auto&& get() { return v; }
        friend std::ostream& operator<<(std::ostream& o, B const& b) {
            o << "B: { ";
            for (auto it = std::begin(b.v); it != std::end(b.v); ++it) { if (it != std::begin(b.v)) { o << " } { "; } o << *it; }
            return o << " }";
        }
    private:
        std::vector<A> v;
};

int main() {
    A a, a1, a2; B b;

    a.get().insert(a.get().end(), {1, 2, 3, 4});
    a1.get().insert(a1.get().end(), {5, 6, 7, 8});
    a2.get().insert(a2.get().end(), {9, 10, 11, 12});

    std::cout << a << "\n" << a1 <<  "\n" << a2;
    b.get().insert(b.get().end(), {a, a1, a2});
    std::cout << "\n" <<  b << "\n";

    a.get().push_back(std::move(b.get()[2].get()[2])); 

    std::cout << a << "\n" << a1 <<  "\n" << a2;
    std::cout << "\n" <<  b;
    return 0;
}

Answer 1

您在std::move上呼叫int。对于非类类型，复制和移动是相同的操作。

该值将被复制而不是被移动。为什么？

1 个答案: