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问题描述

我收到的面试问题如下:

I was given the following as an interview question:

class A
{
public:
    void fun()
    {
        std::cout << "fun" << std::endl;
    }
};

A* a = NULL;
a->fun();

执行此代码时会发生什么,为什么?

What will happen when this code is executed, and why?

推荐答案

这是未定义的行为,所以任何事情都可能发生.

It's undefined behavior, so anything might happen.

一个可能的结果是它只打印 "fun" 因为该方法不访问它被调用的对象的任何成员变量(对象应该居住的内存没有需要访问,所以不一定会发生访问冲突).

A possible result would be that it just prints "fun" since the method doesn't access any member variables of the object it is called on (the memory where the object supposedly lives doesn't need to be accessed, so access violations don't necessarily occur).

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09-14 06:39