您应该会在以下方面出现错误：

  

constexpr Test& Test::operator=(const Test&)’ is implicitly declared as deleted because ‘Test’ declares a move constructor or move assignment operator

这是一件好事，因为默认的副本分配运算符会做得很糟糕。它将复制您的指针-并且您将最终获得两个都声称拥有数据的实例。两者都将稍后尝试delete[]-结果是未定义的行为。

您可能应该实现The rule of three/five/zero中提到的所有5个成员函数。

示例：

#include <algorithm>
#include <iostream>
#include <utility>

class Test {
public:
    int size;
    double* array;
public: 
    Test() : 
        size(0),array(nullptr)                    // make sure the default constructed array
                                          // is set to nullptr
    {}
    Test(int sizeArg) :
        size(sizeArg),array(new double[size]{})         // zero initialize the array
    {}
    // rule of five:
    Test(const Test& arg) :               // copy ctor
        size(arg.size),array(new double[size])
    {
        std::copy(arg.array,arg.array+size,array);
    }
    Test(Test&& arg) :                    // move ctor
        size(arg.size),array(std::exchange(arg.array,nullptr)) // get pointer from arg and give nullptr
                                                 // to arg
    {}
    Test& operator=(const Test& arg) {    // copy assignment operator
        Test tmp(arg);                    // reuse copy constructor
        std::swap(*this,tmp);            // let "tmp" destroy our current array
        return *this;
    } 
    Test& operator=(Test&& arg) {         // move assignment operator 
        size = arg.size;
        std::swap(array,arg.array);      // let the other object delete our current array
        return *this;
    }   
    ~Test() {
        delete[] array;                   // no "if" needed,it's ok to delete nullptr
    }
};

int main()
{
    Test outTest;
    int x = 1;

    if(x==1) {                           // use == when comparing for equality
        Test arg(2);
        outTest = arg;
    }
    std::cout << outTest.array;
}