在分配新对象之前未删除对象

Question

我有点困惑，因为我确定这应该会有所不同。看看这个代码示例：

#include <iostream>
#include <string>

using namespace std;

class base
{
    public:
    virtual ~base() = default;
};

class derived : public base
{
    private:
    int a = 0;
    int *b = nullptr;
    std::string lol;

    public:
    derived(std::string s) : b(new int(6)), lol{s} { cout << "ctor " << lol << endl; }
    derived(derived const& d) : lol{d.lol + " copy"} {cout << "copy " << lol << endl; }

    virtual ~derived() { cout << "dtor " << lol << endl; delete b; }

    virtual void superFunction() { cout << "OMG " << lol << endl; }
};

int main()
{
    derived a("a");
    derived b("b");
    a = b;
}

关闭所有优化后的程序输出是：

ctor a
ctor b
dtor b
dtor b

我确信在这种情况下，编译器应该生成删除对象 a 并使用复制构造函数创建新对象的代码。相反，它使用隐式声明的 operator=。

谁能解释一下为什么？或者指向 C++ 标准。

谢谢。

Answer 1

当你写a = b;时，编译器会调用赋值运算符，如果代码中不存在且未标记为已删除，则会自动生成。仅当您尝试 初始化 来自另一个对象的新对象时才使用复制构造函数，如下所示：

derived a("a");
derived b = a;

此外，您的代码在 main returns 之前崩溃，因为它试图删除 b，它指向 a 和 b 之后的相同内存 a = b; 默认分配。

如果你想在 a = b; 执行后用 derived 析构函数删除 a，你只需要复制和交换惯用语。 What is the copy and swap idiom? has a great answer on how to do that in legacy and modern C++. Proper implementation of rule-of-four from that answer will perfectly fit the DRY principle and help you to avoid memory issues. Note the fabulous trick with passing parameter to operator= by value, which makes compiler select the appropriate constructor (copy or move) and allows you to write only four methods instead of all five 个。

在分配新对象之前未删除对象

Object not deleted before new is assigned

c++

deep-copy

copy-constructor

assignment-operator