这怎么可以编译？ （删除const对象的成员） [英] How can this compile? (delete a member of const object)

问题描述

  class Test $ b $ 
 b {
 public：
 Test（）
 {
p = new int（0）; 
} 
 
 Test（const Test& t）
 {
 delete t.p。 //我期望这里有一个错误
} 
 
〜Test（）
 {
 delete p; 
} 
 
 private：
 int * p; 
}; 
  

解决方案

不改变 p 本身（因为 p 保持不变，因为你不改变它的值）重新改变 p 指向并因此在一个附加级别的间接工作。这是可能的，因为删除与指针相关联的内存不会改变指针本身。

严格意义上，对象的常量被保留，即使 c>

正如JonH提到的那样，它的逻辑常数已经被违反了，在注释中，如果你不能删除由const对象保存的指针指向的对象，那么你将会遇到内存泄漏，因为你无法在对象后正确清理。

I would expect an error inside the copy constructor, but this compiles just fine with MSVC10.

class Test
{
public:
    Test()
    {
        p = new int(0);
    }

    Test(const Test& t)
    {
        delete t.p; // I would expect an error here
    }

    ~Test()
    {
        delete p;
    }

private:
    int* p;
};

解决方案

The issue that you are running into here is that you are not changing p per se (thus pstays immutable as you're not changing its value), but you're changing what p points to and thus are working at one additional level of indirection. This is possible because deleteing the memory associated with a pointer doesn't change the pointer itself.

In a strict sense the const-ness of the object is preserved, even though its logical constness has been violated as you pulled the rug from underneath whatever p was pointing to.

As JonH mentioned in the comment, if you were not able to delete the object pointed to by a pointer held in a const object, you would end up with memory leaks because you wouldn't be able to clean up properly after the object.

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