开发与运维函数问题之自定义拷贝构造函数和拷贝赋值运算符如何解决

#include <iostream>
using namespace std;
class BigMemoryPool {    private:        static const int POOL_SIZE = 4096;        int* mPool;
    public:        BigMemoryPool() : mPool(new int[POOL_SIZE]{0}) {            cout << "call default init" << endl;        }
        // 编译器会优化移动构造函数，正常情况可能不会被执行        // 可以添加编译选项 “-fno-elide-constructors” 关闭优化来观察效果        BigMemoryPool(BigMemoryPool && other) noexcept {            mPool = other.mPool;            other.mPool = nullptr;            cout << "call move init" << endl;        }
        BigMemoryPool & operator=(BigMemoryPool && other) noexcept {            if (this != &other) {                this->mPool = other.mPool;                other.mPool = nullptr;            }            cout << "call op move" << endl;            return *this;        }        void showPoolAddr() {            cout << "pool addr:" << &(mPool[0]) << endl;        }
        ~BigMemoryPool() {            cout << "call destructor" << endl;        }};
BigMemoryPool makeBigMemoryPool() {    BigMemoryPool x;  // 调用默认构造函数    x.showPoolAddr();    return x;         // 返回临时变量，属于右值}
int main() {    BigMemoryPool a(makeBigMemoryPool());      a.showPoolAddr();    a = makeBigMemoryPool();      a.showPoolAddr();    return 0;}
// 输出内容call default initpool addr:0x152009600instance addr:0x16fdfeda0pool addr:0x152009600instance addr:0x16fdfeda0  // 编译器优化，这里a和x其实是同一个实例，因此不会触发移动构造call default initpool addr:0x15200e600    // 新的临时变量，堆内存重新分配instance addr:0x16fdfed88  // 临时变量对象地址call op move        // 移动赋值call destructorpool addr:0x15200e600    // a的Pool指向的内存地址变成新临时对象分配的地址，完成转移instance addr:0x16fdfeda0  // a对象的地址没有变化call destructor