在C#中安全，但在C ++中不安全，简单返回指针/引用

问题描述

C ++代码:

person* NewPerson(void)
{
  person p;
  /* ... */
  return &p; //return pointer to person.
}

C#代码:

person NewPerson()
{
  return new person(); //return reference to person.
}

如果我理解此权利，则C ++中的示例不正确，因为p将超出范围，该函数将返回一个野生指针(悬空指针).

If I understand this right, the example in C++ is not OK, because the p willgo out of scope, and the function will return a wild pointer (dangling pointer).

C#中的示例是可以的，因为只要有匿名新人的引用，它就会一直在作用域内. (调用函数得到一个.)

The example in C# is OK, because the anonymous new person will stay in scope as long as there is a reference to it. (The calling function gets one.)

我说对了吗?

推荐答案

person* NewPerson(void)
{
  person p();
  /* ... */
  return &p; //return pointer to person.
}

p不是person，请参见.这样，您会得到一个编译器错误.

p is not a person, see most vexing parse. As such, you'd get a compiler error.

其余的，是的，你是对的.

For the rest, yes you're right.

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